Optimization in Calculus

Cheatsheet Content

### Introduction to Optimization Problems Optimization problems in differential calculus involve finding the maximum or minimum value of a function, often subject to certain constraints. These problems are ubiquitous in science, engineering, economics, and everyday life, as they help us determine the best possible outcome or most efficient design. The core idea is to use derivatives to locate critical points where the function's slope is zero, which correspond to potential local maxima or minima. #### Steps to Solve Optimization Problems: 1. **Understand the Problem:** Read carefully, identify what needs to be maximized or minimized, and what are the given constraints. 2. **Draw a Diagram (if applicable):** Visualizing the problem often helps in setting up the equations. Label all relevant quantities. 3. **Define Variables:** Assign symbols to all quantities involved. 4. **Formulate the Objective Function:** Write an equation for the quantity to be maximized or minimized (e.g., Area, Volume, Cost, Time) in terms of the defined variables. 5. **Formulate Constraint Equations (if any):** Write equations that relate the variables based on the given constraints. 6. **Reduce to a Single Variable:** Use the constraint equations to express the objective function as a function of a single independent variable. 7. **Find the Domain:** Determine the possible range of values for the independent variable based on the problem's physical or geometric constraints. 8. **Differentiate:** Find the first derivative of the objective function with respect to the single variable. 9. **Find Critical Points:** Set the first derivative equal to zero and solve for the variable. These are potential locations for extrema. Also, consider points where the derivative is undefined or the endpoints of the domain. 10. **Test Critical Points:** Use the First Derivative Test or the Second Derivative Test to determine if each critical point corresponds to a local maximum, local minimum, or neither. * **First Derivative Test:** If $f'(x)$ changes from positive to negative at $c$, then $f(c)$ is a local maximum. If $f'(x)$ changes from negative to positive at $c$, then $f(c)$ is a local minimum. * **Second Derivative Test:** If $f''(c) 0$, then $f(c)$ is a local minimum. If $f''(c) = 0$, the test is inconclusive. 11. **Evaluate Endpoints:** If the domain is a closed interval, evaluate the objective function at the endpoints. 12. **State the Conclusion:** Clearly answer the question posed in the problem, including units. ### Volume Optimization Problems These problems involve maximizing or minimizing the volume of a 3D object, often subject to constraints on surface area or material. #### Example Problem 1: Open-Top Box A rectangular piece of cardboard, 10 inches by 16 inches, has squares cut out from each corner. The flaps are then folded up to form an open-top box. Find the maximum volume of the box. **Solution:** 1. **Diagram:** A rectangle with squares cut from corners. 2. **Variables:** Let $x$ be the side length of the square cut from each corner. * Height of the box: $h = x$ * Length of the box: $l = 16 - 2x$ * Width of the box: $w = 10 - 2x$ 3. **Objective Function:** Volume $V = lwh = (16 - 2x)(10 - 2x)x$ $V(x) = (160 - 32x - 20x + 4x^2)x = (160 - 52x + 4x^2)x = 4x^3 - 52x^2 + 160x$ 4. **Domain:** For the dimensions to be positive, $x > 0$, $16 - 2x > 0 \Rightarrow x 0 \Rightarrow x **Solution:** 1. **Diagram:** A cylinder with radius $r$ and height $h$. 2. **Variables:** $r$ = radius, $h$ = height. 3. **Objective Function:** Minimize surface area $A = 2\pi r^2$ (top and bottom) $+ 2\pi rh$ (side). 4. **Constraint:** Volume $V = \pi r^2 h = 1000$ cm³. 5. **Reduce to Single Variable:** From the constraint, $h = \frac{1000}{\pi r^2}$. Substitute $h$ into the area function: $A(r) = 2\pi r^2 + 2\pi r \left(\frac{1000}{\pi r^2}\right) = 2\pi r^2 + \frac{2000}{r}$ 6. **Domain:** $r > 0$. 7. **Differentiate:** $A'(r) = 4\pi r - \frac{2000}{r^2}$ 8. **Critical Points:** Set $A'(r) = 0$: $4\pi r - \frac{2000}{r^2} = 0 \Rightarrow 4\pi r = \frac{2000}{r^2} \Rightarrow 4\pi r^3 = 2000 \Rightarrow r^3 = \frac{2000}{4\pi} = \frac{500}{\pi}$ $r = \sqrt[3]{\frac{500}{\pi}} \approx 5.419$ cm. 9. **Test Critical Points:** Use the Second Derivative Test: $A''(r) = 4\pi + \frac{4000}{r^3}$ For $r > 0$, $A''(r)$ is always positive, so $A\left(\sqrt[3]{\frac{500}{\pi}}\right)$ is a local minimum. 10. **Dimensions:** $r \approx 5.419$ cm $h = \frac{1000}{\pi (5.419)^2} = \frac{1000}{\pi \left(\frac{500}{\pi}\right)^{2/3}} = \frac{1000}{\pi^{1/3} 500^{2/3}} = \frac{2 \cdot 500}{\pi^{1/3} 500^{2/3}} = \frac{2 \cdot 500^{1/3}}{\pi^{1/3}} = 2 \sqrt[3]{\frac{500}{\pi}} = 2r$ So, $h \approx 2(5.419) = 10.838$ cm. The height should be twice the radius for minimum surface area. #### Example Problem 3: Cone from a Circle A sector of a circle with radius $R$ is cut out and folded to form a cone. What angle of the sector will maximize the volume of the cone? **Solution:** 1. **Diagram:** A circle with a sector missing, and a cone formed. 2. **Variables:** * $R$: radius of the original circle (and slant height of the cone) * $r$: radius of the cone's base * $h$: height of the cone * $\theta$: central angle of the sector (in radians) 3. **Relationships:** * Slant height $L = R$ * Circumference of cone base $2\pi r = R\theta$ (arc length of the sector) $\Rightarrow r = \frac{R\theta}{2\pi}$ * Pythagorean theorem for cone: $r^2 + h^2 = L^2 \Rightarrow r^2 + h^2 = R^2 \Rightarrow h = \sqrt{R^2 - r^2}$ 4. **Objective Function:** Volume of cone $V = \frac{1}{3}\pi r^2 h$ Substitute $r$ and $h$: $V(\theta) = \frac{1}{3}\pi \left(\frac{R\theta}{2\pi}\right)^2 \sqrt{R^2 - \left(\frac{R\theta}{2\pi}\right)^2}$ $V(\theta) = \frac{1}{3}\pi \frac{R^2\theta^2}{4\pi^2} \sqrt{R^2 - \frac{R^2\theta^2}{4\pi^2}} = \frac{R^2\theta^2}{12\pi} \sqrt{R^2 \left(1 - \frac{\theta^2}{4\pi^2}\right)}$ $V(\theta) = \frac{R^2\theta^2}{12\pi} R \sqrt{1 - \frac{\theta^2}{4\pi^2}} = \frac{R^3}{12\pi} \theta^2 \sqrt{1 - \frac{\theta^2}{4\pi^2}}$ To simplify differentiation, maximize $V^2$: $V^2(\theta) = \left(\frac{R^3}{12\pi}\right)^2 \theta^4 \left(1 - \frac{\theta^2}{4\pi^2}\right) = C \left(\theta^4 - \frac{\theta^6}{4\pi^2}\right)$ where $C = \left(\frac{R^3}{12\pi}\right)^2$ 5. **Domain:** $0 0$, we must have $4 - \frac{3\theta^2}{2\pi^2} = 0 \Rightarrow \frac{3\theta^2}{2\pi^2} = 4 \Rightarrow \theta^2 = \frac{8\pi^2}{3}$ $\theta = \sqrt{\frac{8\pi^2}{3}} = \frac{2\pi\sqrt{2}}{\sqrt{3}} = \frac{2\pi\sqrt{6}}{3}$ radians. 8. **Test Critical Points:** $\frac{2\pi\sqrt{6}}{3} \approx \frac{2\pi(2.449)}{3} \approx 5.13$ radians. This is within the domain $0 0$ If $\theta > \frac{2\pi\sqrt{6}}{3}$, e.g., $\theta = 2\pi$: $f'(2\pi) = 4(2\pi)^3 - \frac{3(2\pi)^5}{2\pi^2} = 32\pi^3 - \frac{3 \cdot 32\pi^5}{2\pi^2} = 32\pi^3 - 48\pi^3 = -16\pi^3 **Solution:** 1. **Diagram:** A sphere with an inscribed cylinder. 2. **Variables:** * $R$: radius of the sphere (constant) * $r$: radius of the cylinder's base * $h$: height of the cylinder 3. **Relationships:** A cross-section through the center of the sphere and cylinder forms a rectangle inscribed in a circle. $(r)^2 + \left(\frac{h}{2}\right)^2 = R^2 \Rightarrow r^2 = R^2 - \frac{h^2}{4}$ 4. **Objective Function:** Volume of cylinder $V = \pi r^2 h$ Substitute $r^2$: $V(h) = \pi \left(R^2 - \frac{h^2}{4}\right) h = \pi R^2 h - \frac{\pi}{4} h^3$ 5. **Domain:** $0 0$). 8. **Test Critical Points:** $h = \frac{2\sqrt{3}}{3} R \approx 1.15R$, which is in the domain $(0, 2R)$. Using the Second Derivative Test: $V''(h) = -\frac{6\pi}{4} h = -\frac{3\pi}{2} h$ Since $h > 0$, $V''(h) **Solution:** 1. **Diagram:** A cone with an inscribed cylinder. 2. **Variables:** * $R$: radius of the cone (constant) * $H$: height of the cone (constant) * $r$: radius of the cylinder * $h$: height of the cylinder 3. **Relationships:** Use similar triangles from a cross-section. The large triangle has base $R$ and height $H$. The small triangle (above the cylinder) has base $r$ and height $H-h$. $\frac{r}{R} = \frac{H-h}{H} \Rightarrow r = \frac{R}{H}(H-h)$ or $h = H - \frac{H}{R}r$ 4. **Objective Function:** Volume of cylinder $V = \pi r^2 h$ Substitute $h$: $V(r) = \pi r^2 \left(H - \frac{H}{R}r\right) = \pi H r^2 - \frac{\pi H}{R} r^3$ 5. **Domain:** $0 0$, we have $2 - \frac{3}{R} r = 0 \Rightarrow \frac{3}{R} r = 2 \Rightarrow r = \frac{2}{3} R$. 8. **Test Critical Points:** $r = \frac{2}{3} R$ is in the domain $(0, R)$. Using the Second Derivative Test: $V''(r) = 2\pi H - \frac{6\pi H}{R} r$ $V''\left(\frac{2}{3} R\right) = 2\pi H - \frac{6\pi H}{R} \left(\frac{2}{3} R\right) = 2\pi H - 4\pi H = -2\pi H$ Since $H > 0$, $V''\left(\frac{2}{3} R\right) **Solution:** 1. **Diagram:** An ellipse centered at the origin with an inscribed rectangle. 2. **Variables:** Let $(x, y)$ be the coordinates of the corner of the rectangle in the first quadrant. * Width of the rectangle: $2x$ * Height of the rectangle: $2y$ 3. **Objective Function:** Area $A = (2x)(2y) = 4xy$ 4. **Constraint:** The point $(x, y)$ is on the ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. 5. **Reduce to Single Variable:** From the constraint, $\frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} \Rightarrow y^2 = b^2 \left(1 - \frac{x^2}{a^2}\right) \Rightarrow y = \frac{b}{a}\sqrt{a^2 - x^2}$ (since $y>0$). Substitute $y$ into the area function: $A(x) = 4x \frac{b}{a}\sqrt{a^2 - x^2} = \frac{4b}{a} x \sqrt{a^2 - x^2}$ To simplify differentiation, maximize $A^2$: $A^2(x) = \left(\frac{4b}{a}\right)^2 x^2 (a^2 - x^2) = C (a^2x^2 - x^4)$ where $C = \left(\frac{4b}{a}\right)^2$. 6. **Domain:** $0 0$, $a^2 - 2x^2 = 0 \Rightarrow 2x^2 = a^2 \Rightarrow x^2 = \frac{a^2}{2} \Rightarrow x = \frac{a}{\sqrt{2}} = \frac{a\sqrt{2}}{2}$. 9. **Test Critical Points:** $x = \frac{a\sqrt{2}}{2}$ is in the domain $(0, a)$. Using the Second Derivative Test: $f''(x) = 2a^2 - 12x^2$ $f''\left(\frac{a}{\sqrt{2}}\right) = 2a^2 - 12\left(\frac{a^2}{2}\right) = 2a^2 - 6a^2 = -4a^2 **Solution:** 1. **Diagram:** A right triangle formed by the ladder, ground, and wall. A smaller similar triangle formed by the fence, ground, and ladder. 2. **Variables:** * $L$: length of the ladder (hypotenuse) * $x$: distance from the wall to the base of the ladder * $y$: height on the wall where the ladder touches * Fence height: $5$ ft * Fence distance from wall: $4$ ft 3. **Relationships:** * By Pythagorean theorem: $L^2 = x^2 + y^2 \Rightarrow L = \sqrt{x^2 + y^2}$ * By similar triangles: $\frac{y}{x} = \frac{5}{x-4}$ (assuming $x > 4$) $\Rightarrow y = \frac{5x}{x-4}$ 4. **Objective Function:** Minimize $L$. It's easier to minimize $L^2$. $L^2(x) = x^2 + \left(\frac{5x}{x-4}\right)^2 = x^2 + \frac{25x^2}{(x-4)^2}$ 5. **Domain:** $x > 4$. 6. **Differentiate:** Let $f(x) = x^2 + 25x^2(x-4)^{-2}$ $f'(x) = 2x + 50x(x-4)^{-2} - 50x^2(x-4)^{-3}$ $f'(x) = 2x + \frac{50x}{(x-4)^2} - \frac{50x^2}{(x-4)^3}$ $f'(x) = 2x + \frac{50x(x-4) - 50x^2}{(x-4)^3} = 2x + \frac{50x^2 - 200x - 50x^2}{(x-4)^3} = 2x - \frac{200x}{(x-4)^3}$ 7. **Critical Points:** Set $f'(x) = 0$: $2x - \frac{200x}{(x-4)^3} = 0$ $2x \left(1 - \frac{100}{(x-4)^3}\right) = 0$ Since $x > 0$, we must have $1 - \frac{100}{(x-4)^3} = 0 \Rightarrow (x-4)^3 = 100$ $x-4 = \sqrt[3]{100} \Rightarrow x = 4 + \sqrt[3]{100}$ $\sqrt[3]{100} \approx 4.64$, so $x \approx 8.64$ ft. This is in the domain $x > 4$. 8. **Test Critical Points:** If $x 4 + \sqrt[3]{100}$, e.g., $x=10$: $f'(10) = 2(10) - \frac{200(10)}{(10-4)^3} = 20 - \frac{2000}{216} = 20 - 9.26 > 0$ Since $f'(x)$ changes from negative to positive, this is a local minimum. 9. **Shortest Length:** $x = 4 + \sqrt[3]{100}$ $y = \frac{5(4 + \sqrt[3]{100})}{\sqrt[3]{100}} = \frac{5 \cdot 4}{\sqrt[3]{100}} + 5 = \frac{20}{\sqrt[3]{100}} + 5$ $L = \sqrt{x^2 + y^2} = \sqrt{(4 + \sqrt[3]{100})^2 + \left(\frac{5(4 + \sqrt[3]{100})}{\sqrt[3]{100}}\right)^2}$ A simpler approach: from $1 - \frac{100}{(x-4)^3} = 0$, we have $(x-4)^3 = 100$. Also, $y = \frac{5x}{x-4}$. Substitute $x-4 = 100^{1/3}$ into $y$: $y = \frac{5(4+100^{1/3})}{100^{1/3}} = 5\left(\frac{4}{100^{1/3}} + 1\right) = 5\left(4 \cdot 100^{-1/3} + 1\right)$. From similar triangles, $y/x = 5/(x-4)$, so $y(x-4) = 5x$. $L = \sqrt{x^2+y^2}$. Alternative form for $L$: $L = \frac{x}{\cos \alpha} = \frac{y}{\sin \alpha}$ where $\alpha$ is angle with ground. From $\frac{y}{x} = \frac{5}{x-4}$, let $m = \frac{y}{x}$ be the slope. $L^2 = x^2 + y^2$. At the critical point $x-4 = 100^{1/3}$. $2x = \frac{200x}{(x-4)^3} \Rightarrow (x-4)^3 = 100$. $y = \frac{5x}{x-4}$. $L = \sqrt{x^2 + y^2} = \sqrt{x^2 + \left(\frac{5x}{100^{1/3}}\right)^2} = x \sqrt{1 + \frac{25}{100^{2/3}}}$. $x = 4 + 100^{1/3}$. $L = (4 + 100^{1/3}) \sqrt{1 + \frac{25}{(100^{1/3})^2}} = (4 + 100^{1/3}) \sqrt{1 + \frac{25}{100^{2/3}}}$ Let $k = 100^{1/3}$. $L = (4+k)\sqrt{1 + \frac{25}{k^2}}$. $L = \sqrt{(4+k)^2 (1 + \frac{25}{k^2})} = \sqrt{(16 + 8k + k^2) \frac{k^2+25}{k^2}}$ This is getting complicated. Let's use the power of the critical point $x=4+100^{1/3}$. $y = \frac{5(4+100^{1/3})}{100^{1/3}} = 5\left(\frac{4}{100^{1/3}}+1\right) = 5(4 \cdot 100^{-1/3} + 1)$. $L^2 = (4+100^{1/3})^2 + 25(1 + 4 \cdot 100^{-1/3})^2$. This is numerically $L \approx 19.98$ ft. A known result for this problem is $L = (a^{2/3} + b^{2/3})^{3/2}$ where $a$ is fence height, $b$ is fence distance. Here $a=5$, $b=4$. $L = (5^{2/3} + 4^{2/3})^{3/2} = (2.924 + 2.520)^{3/2} = (5.444)^{3/2} \approx 12.72$ ft. Let's recheck the derivative. $f'(x) = 2x - \frac{200x}{(x-4)^3} = 2x \left(1 - \frac{100}{(x-4)^3}\right)$. This is correct. So $x = 4 + 100^{1/3}$. $y = \frac{5x}{x-4} = \frac{5(4+100^{1/3})}{100^{1/3}} = 5\left(1 + \frac{4}{100^{1/3}}\right)$. $L^2 = x^2+y^2 = (4+100^{1/3})^2 + 25\left(1+\frac{4}{100^{1/3}}\right)^2$. This calculation is correct. The result $L = (a^{2/3} + b^{2/3})^{3/2}$ is for the distance from the origin to the point $(x,y)$ on the line tangent to $(b,a)$ where the line is $y-a = m(x-b)$. No, this is for a line segment of length $L$ touching axes at $(x,0)$ and $(0,y)$, passing through $(b,a)$. Here, $x$ is the base, $y$ is the height. The fence is at $(4,5)$. The line equation is $\frac{X}{x} + \frac{Y}{y} = 1$. Since $(4,5)$ is on the line: $\frac{4}{x} + \frac{5}{y} = 1$. From this, $\frac{5}{y} = 1 - \frac{4}{x} = \frac{x-4}{x} \Rightarrow y = \frac{5x}{x-4}$. This is correct. So the calculation of $x$ is also correct. $L = \sqrt{x^2+y^2} = \sqrt{(4+100^{1/3})^2 + \left(\frac{5(4+100^{1/3})}{100^{1/3}}\right)^2}$ Let $k=100^{1/3}$. $L = \sqrt{(4+k)^2 + \frac{25(4+k)^2}{k^2}} = \sqrt{(4+k)^2 \left(1 + \frac{25}{k^2}\right)}$ $L = (4+k) \sqrt{1 + \frac{25}{k^2}} = (4+k) \sqrt{\frac{k^2+25}{k^2}} = (4+k) \frac{\sqrt{k^2+25}}{k}$ $L = \left(\frac{4}{k} + 1\right) \sqrt{k^2+25}$ $k = 100^{1/3}$. $k^2 = 100^{2/3}$. $L = \left(\frac{4}{100^{1/3}} + 1\right) \sqrt{100^{2/3}+25}$. This is the expression for the shortest length. Numerically, $L \approx 12.72$ ft. #### Example Problem 8: Minimum Surface Area of Aquarium An aquarium with a square base and open top is to be constructed from glass. The aquarium must hold 32 cubic feet of water. Find the dimensions that will minimize the amount of glass used. **Solution:** 1. **Diagram:** An open-top box with square base. 2. **Variables:** Let $s$ be the side length of the square base, and $h$ be the height. 3. **Objective Function:** Minimize surface area $A = s^2$ (base) $+ 4sh$ (four sides). 4. **Constraint:** Volume $V = s^2 h = 32$ cubic feet. 5. **Reduce to Single Variable:** From the constraint, $h = \frac{32}{s^2}$. Substitute $h$ into the area function: $A(s) = s^2 + 4s \left(\frac{32}{s^2}\right) = s^2 + \frac{128}{s}$ 6. **Domain:** $s > 0$. 7. **Differentiate:** $A'(s) = 2s - \frac{128}{s^2}$ 8. **Critical Points:** Set $A'(s) = 0$: $2s - \frac{128}{s^2} = 0 \Rightarrow 2s = \frac{128}{s^2} \Rightarrow 2s^3 = 128 \Rightarrow s^3 = 64 \Rightarrow s = 4$. 9. **Test Critical Points:** $s=4$ is in the domain $s > 0$. Using the Second Derivative Test: $A''(s) = 2 + \frac{256}{s^3}$ $A''(4) = 2 + \frac{256}{4^3} = 2 + \frac{256}{64} = 2 + 4 = 6 > 0$. This indicates a local minimum. 10. **Dimensions:** $s = 4$ feet $h = \frac{32}{4^2} = \frac{32}{16} = 2$ feet The dimensions are 4 ft x 4 ft x 2 ft. Minimum surface area: $A(4) = 4^2 + \frac{128}{4} = 16 + 32 = 48$ square feet. #### Example Problem 9: Maximize Area of Rectangle in Parabola Find the area of the largest rectangle that can be inscribed in the first quadrant under the parabola $y = 12 - x^2$. **Solution:** 1. **Diagram:** A parabola opening downwards, intersecting the x-axis at $x=\sqrt{12}$ and the y-axis at $y=12$. A rectangle with one corner at the origin and the opposite corner on the parabola. 2. **Variables:** Let $(x, y)$ be the coordinates of the upper-right corner of the rectangle. * Width of the rectangle: $x$ * Height of the rectangle: $y$ 3. **Objective Function:** Area $A = xy$ 4. **Constraint:** The point $(x, y)$ is on the parabola $y = 12 - x^2$. 5. **Reduce to Single Variable:** Substitute $y$ into the area function: $A(x) = x(12 - x^2) = 12x - x^3$ 6. **Domain:** For the rectangle to be in the first quadrant, $x > 0$ and $y > 0$. $12 - x^2 > 0 \Rightarrow x^2 0$). 9. **Test Critical Points:** $x=2$ is in the domain $(0, 2\sqrt{3})$. Using the Second Derivative Test: $A''(x) = -6x$ $A''(2) = -6(2) = -12 **Solution:** 1. **Diagram:** The curve $y=\sqrt{x}$ (top half of a parabola) and the point $(3,0)$. 2. **Variables:** Let $(x, y)$ be a point on the curve. 3. **Objective Function:** Minimize the distance $D$ between $(x, y)$ and $(3, 0)$. $D = \sqrt{(x-3)^2 + (y-0)^2} = \sqrt{(x-3)^2 + y^2}$ It's easier to minimize $D^2$. Let $f(x,y) = D^2 = (x-3)^2 + y^2$. 4. **Constraint:** The point $(x, y)$ is on the curve $y = \sqrt{x}$. 5. **Reduce to Single Variable:** Substitute $y = \sqrt{x}$ into $f(x,y)$: $f(x) = (x-3)^2 + (\sqrt{x})^2 = (x-3)^2 + x = x^2 - 6x + 9 + x = x^2 - 5x + 9$ 6. **Domain:** For $y=\sqrt{x}$ to be defined, $x \ge 0$. 7. **Differentiate:** $f'(x) = 2x - 5$ 8. **Critical Points:** Set $f'(x) = 0$: $2x - 5 = 0 \Rightarrow x = \frac{5}{2}$. 9. **Test Critical Points:** $x = \frac{5}{2}$ is in the domain $x \ge 0$. Using the Second Derivative Test: $f''(x) = 2$. $f''\left(\frac{5}{2}\right) = 2 > 0$. This indicates a local minimum. 10. **Closest Point:** $x = \frac{5}{2}$ $y = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}$ The closest point on the curve is $\left(\frac{5}{2}, \frac{\sqrt{10}}{2}\right)$. Minimum distance squared: $f\left(\frac{5}{2}\right) = \left(\frac{5}{2}\right)^2 - 5\left(\frac{5}{2}\right) + 9 = \frac{25}{4} - \frac{25}{2} + 9 = \frac{25 - 50 + 36}{4} = \frac{11}{4}$. Minimum distance: $D = \sqrt{\frac{11}{4}} = \frac{\sqrt{11}}{2}$. ### Geometry Optimization Problems These problems focus on optimizing geometric quantities like area, perimeter, or specific dimensions. #### Example Problem 1: Maximize Area of Rectangle with Fixed Perimeter Find the dimensions of a rectangle with perimeter 100 meters that has the largest possible area. **Solution:** 1. **Diagram:** A rectangle. 2. **Variables:** Let $L$ be the length and $W$ be the width of the rectangle. 3. **Objective Function:** Maximize Area $A = LW$. 4. **Constraint:** Perimeter $P = 2L + 2W = 100$. 5. **Reduce to Single Variable:** From the constraint, $2W = 100 - 2L \Rightarrow W = 50 - L$. Substitute $W$ into the area function: $A(L) = L(50 - L) = 50L - L^2$. 6. **Domain:** For a rectangle, $L > 0$ and $W > 0$. $50 - L > 0 \Rightarrow L **Solution:** 1. **Diagram:** A rectangle with one side on the river. 2. **Variables:** Let $L$ be the length parallel to the river, and $W$ be the width perpendicular to the river. 3. **Objective Function:** Maximize Area $A = LW$. 4. **Constraint:** Total fencing $P = L + 2W = 2400$. 5. **Reduce to Single Variable:** From the constraint, $L = 2400 - 2W$. Substitute $L$ into the area function: $A(W) = (2400 - 2W)W = 2400W - 2W^2$. 6. **Domain:** For a field, $W > 0$ and $L > 0$. $2400 - 2W > 0 \Rightarrow 2400 > 2W \Rightarrow W **Solution:** 1. **Diagram:** A light source above the x-axis. 2. **Variables:** $h$: height of light source, $x_0$: fixed distance of object. 3. **Objective Function:** Maximize $I(h) = k h (h^2+x_0^2)^{-3/2}$. (Assume $k>0$). 4. **Domain:** $h > 0$. 5. **Differentiate:** Use the product rule. $I'(h) = k \left[ 1 \cdot (h^2+x_0^2)^{-3/2} + h \cdot \left(-\frac{3}{2}\right)(h^2+x_0^2)^{-5/2} \cdot (2h) \right]$ $I'(h) = k (h^2+x_0^2)^{-5/2} \left[ (h^2+x_0^2) - 3h^2 \right]$ $I'(h) = k (h^2+x_0^2)^{-5/2} (x_0^2 - 2h^2)$ 6. **Critical Points:** Set $I'(h) = 0$. Since $k > 0$ and $(h^2+x_0^2)^{-5/2} > 0$, we must have: $x_0^2 - 2h^2 = 0 \Rightarrow x_0^2 = 2h^2 \Rightarrow h^2 = \frac{x_0^2}{2} \Rightarrow h = \frac{x_0}{\sqrt{2}} = \frac{x_0\sqrt{2}}{2}$ (since $h>0$). 7. **Test Critical Points:** If $h 0$, so $I'(h) > 0$. If $h > \frac{x_0}{\sqrt{2}}$, e.g., $h$ very large: $x_0^2 - 2h^2 **Solution:** 1. **Diagram:** Coordinate plane with intersection at origin $(0,0)$. Car A moves along x-axis, Car B along y-axis. 2. **Variables:** * Time $t$ (in hours) after $t=0$. * Position of Car A: $(x_A(t), 0)$. Initial $x_A(0) = 100$. Speed $50$ mph west, so $x_A(t) = 100 - 50t$. * Position of Car B: $(0, y_B(t))$. Initial $y_B(0) = -90$. Speed $60$ mph north, so $y_B(t) = -90 + 60t$. 3. **Objective Function:** Minimize the distance $D$ between Car A and Car B. $D(t) = \sqrt{(x_A(t) - 0)^2 + (0 - y_B(t))^2} = \sqrt{(100 - 50t)^2 + (-(-90 + 60t))^2}$ $D(t) = \sqrt{(100 - 50t)^2 + (90 - 60t)^2}$ It's easier to minimize $D^2$. Let $f(t) = (100 - 50t)^2 + (90 - 60t)^2$. 4. **Domain:** $t \ge 0$. Car A reaches intersection at $t=100/50=2$. Car B reaches intersection at $t=90/60=1.5$. So the domain is relevant for $t \ge 0$. 5. **Differentiate:** $f'(t) = 2(100 - 50t)(-50) + 2(90 - 60t)(-60)$ $f'(t) = -100(100 - 50t) - 120(90 - 60t)$ $f'(t) = -10000 + 5000t - 10800 + 7200t$ $f'(t) = 12200t - 20800$ 6. **Critical Points:** Set $f'(t) = 0$: $12200t - 20800 = 0 \Rightarrow 12200t = 20800 \Rightarrow t = \frac{20800}{12200} = \frac{208}{122} = \frac{104}{61} \approx 1.705$ hours. 7. **Test Critical Points:** $t \approx 1.705$ is in the domain $t \ge 0$. Using the Second Derivative Test: $f''(t) = 12200$. $f''(t) = 12200 > 0$. This indicates a local minimum. 8. **Conclusion:** The cars are closest at $t = \frac{104}{61}$ hours (approximately 1 hour, 42 minutes, 18 seconds). Minimum distance: $x_A\left(\frac{104}{61}\right) = 100 - 50\left(\frac{104}{61}\right) = \frac{6100 - 5200}{61} = \frac{900}{61}$ $y_B\left(\frac{104}{61}\right) = -90 + 60\left(\frac{104}{61}\right) = \frac{-5490 + 6240}{61} = \frac{750}{61}$ $D = \sqrt{\left(\frac{900}{61}\right)^2 + \left(\frac{750}{61}\right)^2} = \sqrt{\frac{810000 + 562500}{61^2}} = \frac{\sqrt{1372500}}{61} = \frac{\sqrt{25 \cdot 54900}}{61} = \frac{5 \sqrt{54900}}{61} = \frac{5 \sqrt{9 \cdot 6100}}{61} = \frac{5 \cdot 3 \sqrt{6100}}{61} = \frac{15 \sqrt{100 \cdot 61}}{61} = \frac{15 \cdot 10 \sqrt{61}}{61} = \frac{150\sqrt{61}}{61}$ miles. $D \approx \frac{150 \cdot 7.81}{61} \approx 19.2$ miles. #### Example Problem 5: Optimal Point on a Line Find the point on the line $y = 2x - 3$ that is closest to the origin $(0,0)$. **Solution:** 1. **Diagram:** A line and the origin. 2. **Variables:** Let $(x, y)$ be a point on the line. 3. **Objective Function:** Minimize the distance $D$ from $(x, y)$ to $(0,0)$. $D = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}$. Minimize $D^2 = f(x,y) = x^2 + y^2$. 4. **Constraint:** The point $(x, y)$ lies on the line $y = 2x - 3$. 5. **Reduce to Single Variable:** Substitute $y = 2x - 3$ into $f(x,y)$: $f(x) = x^2 + (2x - 3)^2 = x^2 + (4x^2 - 12x + 9) = 5x^2 - 12x + 9$. 6. **Domain:** All real numbers for $x$. 7. **Differentiate:** $f'(x) = 10x - 12$. 8. **Critical Points:** Set $f'(x) = 0$: $10x - 12 = 0 \Rightarrow 10x = 12 \Rightarrow x = \frac{12}{10} = \frac{6}{5}$. 9. **Test Critical Points:** Using the Second Derivative Test: $f''(x) = 10$. $f''\left(\frac{6}{5}\right) = 10 > 0$. This indicates a local minimum. 10. **Closest Point:** $x = \frac{6}{5}$ $y = 2\left(\frac{6}{5}\right) - 3 = \frac{12}{5} - \frac{15}{5} = -\frac{3}{5}$. The closest point on the line is $\left(\frac{6}{5}, -\frac{3}{5}\right)$. Minimum distance squared: $f\left(\frac{6}{5}\right) = 5\left(\frac{6}{5}\right)^2 - 12\left(\frac{6}{5}\right) + 9 = 5\left(\frac{36}{25}\right) - \frac{72}{5} + 9 = \frac{36}{5} - \frac{72}{5} + \frac{45}{5} = \frac{36 - 72 + 45}{5} = \frac{9}{5}$. Minimum distance: $D = \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5}$. #### Example Problem 6: Maximize Area of Sector A sector of a circle has a perimeter of 100 cm. Find the radius and central angle that maximize its area. **Solution:** 1. **Diagram:** A sector of a circle. 2. **Variables:** $r$: radius, $s$: arc length, $\theta$: central angle (in radians). 3. **Objective Function:** Maximize Area $A = \frac{1}{2}r^2\theta$. 4. **Constraints:** * Perimeter $P = 2r + s = 100$. * Relationship between arc length and angle: $s = r\theta$. 5. **Reduce to Single Variable:** From $P = 2r + s$, we have $s = 100 - 2r$. Substitute $s$ into $s=r\theta$: $100 - 2r = r\theta \Rightarrow \theta = \frac{100 - 2r}{r} = \frac{100}{r} - 2$. Substitute $\theta$ into the area function: $A(r) = \frac{1}{2}r^2 \left(\frac{100}{r} - 2\right) = \frac{1}{2}r^2 \frac{100 - 2r}{r} = \frac{1}{2}r(100 - 2r) = 50r - r^2$. 6. **Domain:** $r > 0$. Also, $\theta > 0 \Rightarrow \frac{100}{r} - 2 > 0 \Rightarrow \frac{100}{r} > 2 \Rightarrow 100 > 2r \Rightarrow r **Solution:** 1. **Diagram:** A semicircle with an inscribed rectangle. The base of the rectangle lies on the diameter of the semicircle. 2. **Variables:** Let $(x, y)$ be the coordinates of the upper-right corner of the rectangle. * Width of the rectangle: $2x$ * Height of the rectangle: $y$ 3. **Objective Function:** Area $A = (2x)y = 2xy$. 4. **Constraint:** The point $(x, y)$ is on the semicircle $x^2 + y^2 = R^2$ (for $y \ge 0$). 5. **Reduce to Single Variable:** From the constraint, $y = \sqrt{R^2 - x^2}$. Substitute $y$ into the area function: $A(x) = 2x\sqrt{R^2 - x^2}$. To simplify differentiation, maximize $A^2$: $A^2(x) = (2x)^2 (R^2 - x^2) = 4x^2(R^2 - x^2) = 4R^2x^2 - 4x^4$. 6. **Domain:** $0 0$, $R^2 - 2x^2 = 0 \Rightarrow 2x^2 = R^2 \Rightarrow x^2 = \frac{R^2}{2} \Rightarrow x = \frac{R}{\sqrt{2}} = \frac{R\sqrt{2}}{2}$. 9. **Test Critical Points:** $x = \frac{R\sqrt{2}}{2}$ is in the domain $(0, R)$. Using the Second Derivative Test: $f''(x) = 8R^2 - 48x^2$. $f''\left(\frac{R}{\sqrt{2}}\right) = 8R^2 - 48\left(\frac{R^2}{2}\right) = 8R^2 - 24R^2 = -16R^2 **Solution:** 1. **Diagram:** Two points and a point on the x-axis. 2. **Variables:** Let the pumping station be at $(x, 0)$. * Factory 1: $F_1(0,1)$ * Factory 2: $F_2(3,5)$ 3. **Objective Function:** Minimize total length $L = D_1 + D_2$. $D_1 = \sqrt{(x-0)^2 + (0-1)^2} = \sqrt{x^2 + 1}$ $D_2 = \sqrt{(x-3)^2 + (0-5)^2} = \sqrt{(x-3)^2 + 25}$ $L(x) = \sqrt{x^2 + 1} + \sqrt{(x-3)^2 + 25}$. 4. **Domain:** All real numbers for $x$. 5. **Differentiate:** $L'(x) = \frac{1}{2\sqrt{x^2+1}}(2x) + \frac{1}{2\sqrt{(x-3)^2+25}}(2(x-3))$ $L'(x) = \frac{x}{\sqrt{x^2+1}} + \frac{x-3}{\sqrt{(x-3)^2+25}}$ 6. **Critical Points:** Set $L'(x) = 0$: $\frac{x}{\sqrt{x^2+1}} = -\frac{x-3}{\sqrt{(x-3)^2+25}}$ This can be interpreted geometrically. The expression $\frac{x}{\sqrt{x^2+1}}$ is $\cos \alpha$ where $\alpha$ is the angle the first pipe makes with the negative x-axis. Similarly for the second pipe. This implies the angle of incidence equals the angle of reflection (Fermat's principle of least time). Let $\frac{x}{\sqrt{x^2+1}} = \frac{3-x}{\sqrt{(x-3)^2+25}}$. Squaring both sides is often done but can introduce extraneous solutions. A more robust approach: the derivative is $\frac{x}{\sqrt{x^2+1}} + \frac{x-3}{\sqrt{(x-3)^2+25}}$. Let's consider the slopes. The slope of the line from $(0,1)$ to $(x,0)$ is $\frac{0-1}{x-0} = -\frac{1}{x}$. The slope of the line from $(3,5)$ to $(x,0)$ is $\frac{0-5}{x-3} = -\frac{5}{x-3}$. The condition for minimum distance (by Fermat's Principle / reflection) is that the angle of incidence equals the angle of reflection. This means if we reflect one point across the x-axis, the shortest path is a straight line to the other point. Reflect $F_1(0,1)$ to $F_1'(0,-1)$. The distance from $F_1$ to $(x,0)$ and $F_1'$ to $(x,0)$ is the same. Now we need to find the point $(x,0)$ that minimizes the distance between $F_1'(0,-1)$ and $F_2(3,5)$. This is the straight line connecting them. The equation of the line connecting $(0,-1)$ and $(3,5)$: Slope $m = \frac{5 - (-1)}{3 - 0} = \frac{6}{3} = 2$. Equation: $y - (-1) = 2(x - 0) \Rightarrow y + 1 = 2x \Rightarrow y = 2x - 1$. The point on the x-axis is when $y=0$: $0 = 2x - 1 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}$. 7. **Critical Point:** $x = \frac{1}{2}$. Now, verify with the derivative: $L'\left(\frac{1}{2}\right) = \frac{1/2}{\sqrt{(1/2)^2+1}} + \frac{1/2-3}{\sqrt{(1/2-3)^2+25}} = \frac{1/2}{\sqrt{1/4+1}} + \frac{-5/2}{\sqrt{25/4+25}}$ $= \frac{1/2}{\sqrt{5/4}} + \frac{-5/2}{\sqrt{125/4}} = \frac{1/2}{\sqrt{5}/2} + \frac{-5/2}{5\sqrt{5}/2} = \frac{1}{\sqrt{5}} - \frac{1}{\sqrt{5}} = 0$. So $x = \frac{1}{2}$ is indeed a critical point. 8. **Test Critical Points:** The function $L(x)$ is convex (sum of convex functions), so this critical point is a global minimum. 9. **Conclusion:** The pumping station should be located at $\left(\frac{1}{2}, 0\right)$. Minimum total length: $L\left(\frac{1}{2}\right) = \sqrt{\left(\frac{1}{2}\right)^2 + 1} + \sqrt{\left(\frac{1}{2}-3\right)^2 + 25} = \sqrt{\frac{1}{4} + 1} + \sqrt{\left(-\frac{5}{2}\right)^2 + 25}$ $= \sqrt{\frac{5}{4}} + \sqrt{\frac{25}{4} + 25} = \frac{\sqrt{5}}{2} + \sqrt{\frac{25+100}{4}} = \frac{\sqrt{5}}{2} + \sqrt{\frac{125}{4}} = \frac{\sqrt{5}}{2} + \frac{5\sqrt{5}}{2} = \frac{6\sqrt{5}}{2} = 3\sqrt{5}$. #### Example Problem 9: Angle for Maximum Range A projectile is fired from the origin with initial velocity $v_0$ at an angle $\theta$ with the horizontal. Its range $R$ (horizontal distance) is given by $R = \frac{v_0^2 \sin(2\theta)}{g}$, where $g$ is the acceleration due to gravity. Find the angle $\theta$ that maximizes the range. **Solution:** 1. **Diagram:** A projectile trajectory. 2. **Variables:** $\theta$: launch angle, $v_0$: initial velocity (constant), $g$: gravity (constant). 3. **Objective Function:** Maximize $R(\theta) = \frac{v_0^2}{g} \sin(2\theta)$. 4. **Domain:** Physically, $0 \le \theta \le \frac{\pi}{2}$ (0 to 90 degrees). 5. **Differentiate:** $R'(\theta) = \frac{v_0^2}{g} \cdot \cos(2\theta) \cdot 2 = \frac{2v_0^2}{g} \cos(2\theta)$. 6. **Critical Points:** Set $R'(\theta) = 0$: $\frac{2v_0^2}{g} \cos(2\theta) = 0$. Since $\frac{2v_0^2}{g} \ne 0$, we need $\cos(2\theta) = 0$. For $0 \le \theta \le \frac{\pi}{2}$, we have $0 \le 2\theta \le \pi$. In this interval, $\cos(2\theta) = 0$ when $2\theta = \frac{\pi}{2}$. So, $\theta = \frac{\pi}{4}$. 7. **Test Critical Points:** Using the Second Derivative Test: $R''(\theta) = \frac{2v_0^2}{g} (-\sin(2\theta) \cdot 2) = -\frac{4v_0^2}{g} \sin(2\theta)$. $R''\left(\frac{\pi}{4}\right) = -\frac{4v_0^2}{g} \sin\left(2 \cdot \frac{\pi}{4}\right) = -\frac{4v_0^2}{g} \sin\left(\frac{\pi}{2}\right) = -\frac{4v_0^2}{g} (1) = -\frac{4v_0^2}{g}$. Since $v_0^2/g > 0$, $R''\left(\frac{\pi}{4}\right) **Solution:** 1. **Diagram:** A cylinder with radius $r$ and height $h$. 2. **Variables:** $r$ = radius, $h$ = height. 3. **Objective Function:** Minimize surface area $A = 2\pi r^2$ (top and bottom) $+ 2\pi rh$ (side). 4. **Constraint:** Volume $V = \pi r^2 h = 100$ cm³. 5. **Reduce to Single Variable:** From the constraint, $h = \frac{100}{\pi r^2}$. Substitute $h$ into the area function: $A(r) = 2\pi r^2 + 2\pi r \left(\frac{100}{\pi r^2}\right) = 2\pi r^2 + \frac{200}{r}$. 6. **Domain:** $r > 0$. 7. **Differentiate:** $A'(r) = 4\pi r - \frac{200}{r^2}$. 8. **Critical Points:** Set $A'(r) = 0$: $4\pi r - \frac{200}{r^2} = 0 \Rightarrow 4\pi r = \frac{200}{r^2} \Rightarrow 4\pi r^3 = 200 \Rightarrow r^3 = \frac{200}{4\pi} = \frac{50}{\pi}$. $r = \sqrt[3]{\frac{50}{\pi}} \approx 2.515$ cm. 9. **Test Critical Points:** Use the Second Derivative Test: $A''(r) = 4\pi + \frac{400}{r^3}$. For $r > 0$, $A''(r)$ is always positive, so $A\left(\sqrt[3]{\frac{50}{\pi}}\right)$ is a local minimum. 10. **Dimensions:** $r = \sqrt[3]{\frac{50}{\pi}}$ cm $h = \frac{100}{\pi \left(\sqrt[3]{\frac{50}{\pi}}\right)^2} = \frac{100}{\pi \left(\frac{50}{\pi}\right)^{2/3}} = \frac{100}{\pi^{1/3} 50^{2/3}} = \frac{2 \cdot 50}{\pi^{1/3} 50^{2/3}} = \frac{2 \cdot 50^{1/3}}{\pi^{1/3}} = 2 \sqrt[3]{\frac{50}{\pi}} = 2r$. So, $h = 2r$. Minimum surface area: $A = 2\pi r^2 + 2\pi r (2r) = 2\pi r^2 + 4\pi r^2 = 6\pi r^2 = 6\pi \left(\frac{50}{\pi}\right)^{2/3} \approx 119.6$ cm². ### Time Rates Optimization Problems These problems involve finding the optimal time or rate to achieve a certain goal, often related to cost, speed, or efficiency. While related to related rates, optimization here focuses on finding an extremum of a function involving time. #### Example Problem 1: Optimal Travel Route (Refraction) A man is in a boat 2 miles from the nearest point on a straight shoreline. He wishes to reach a house 6 miles down the shore. He can row at 2 mph and walk at 4 mph. Where should he land the boat to reach the house in the shortest possible time? **Solution:** 1. **Diagram:** A right triangle (boat to shore point) and a straight line along the shore. 2. **Variables:** * Let the boat's starting point be $(0, 2)$. * Let the nearest point on shore be $(0, 0)$. * Let the house be at $(6, 0)$. * Let the landing point be $(x, 0)$. 3. **Objective Function:** Minimize total time $T = T_{\text{row}} + T_{\text{walk}}$. * Distance rowed: $D_{\text{row}} = \sqrt{(x-0)^2 + (0-2)^2} = \sqrt{x^2 + 4}$. * Time rowed: $T_{\text{row}} = \frac{D_{\text{row}}}{\text{row speed}} = \frac{\sqrt{x^2 + 4}}{2}$. * Distance walked: $D_{\text{walk}} = 6 - x$. * Time walked: $T_{\text{walk}} = \frac{D_{\text{walk}}}{\text{walk speed}} = \frac{6 - x}{4}$. $T(x) = \frac{\sqrt{x^2 + 4}}{2} + \frac{6 - x}{4}$. 4. **Domain:** The landing point $x$ must be between $0$ (landing directly opposite) and $6$ (rowing directly to house). So, $0 \le x \le 6$. 5. **Differentiate:** $T'(x) = \frac{1}{2} \cdot \frac{1}{2\sqrt{x^2+4}}(2x) + \frac{1}{4}(-1)$ $T'(x) = \frac{x}{2\sqrt{x^2+4}} - \frac{1}{4}$. 6. **Critical Points:** Set $T'(x) = 0$: $\frac{x}{2\sqrt{x^2+4}} - \frac{1}{4} = 0 \Rightarrow \frac{x}{2\sqrt{x^2+4}} = \frac{1}{4} \Rightarrow 4x = 2\sqrt{x^2+4}$ $2x = \sqrt{x^2+4}$. Square both sides: $(2x)^2 = x^2+4 \Rightarrow 4x^2 = x^2+4 \Rightarrow 3x^2 = 4 \Rightarrow x^2 = \frac{4}{3}$. $x = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \approx 1.155$ miles (since $x \ge 0$). 7. **Test Critical Points:** $x = \frac{2\sqrt{3}}{3}$ is in the domain $[0, 6]$. Using the Second Derivative Test (optional, but good practice): $T''(x) = \frac{1}{2} \frac{\sqrt{x^2+4}(1) - x \cdot \frac{1}{2\sqrt{x^2+4}}(2x)}{x^2+4} = \frac{1}{2} \frac{(x^2+4) - x^2}{(x^2+4)^{3/2}} = \frac{1}{2} \frac{4}{(x^2+4)^{3/2}} = \frac{2}{(x^2+4)^{3/2}}$. Since $T''(x) > 0$ for all $x$, the critical point is a local minimum. Also check endpoints: $T(0) = \frac{\sqrt{0^2+4}}{2} + \frac{6-0}{4} = \frac{2}{2} + \frac{6}{4} = 1 + 1.5 = 2.5$ hours. $T(6) = \frac{\sqrt{6^2+4}}{2} + \frac{6-6}{4} = \frac{\sqrt{36+4}}{2} + 0 = \frac{\sqrt{40}}{2} = \frac{2\sqrt{10}}{2} = \sqrt{10} \approx 3.16$ hours. $T\left(\frac{2\sqrt{3}}{3}\right) = \frac{\sqrt{(2\sqrt{3}/3)^2+4}}{2} + \frac{6 - 2\sqrt{3}/3}{4} = \frac{\sqrt{4/3+4}}{2} + \frac{6 - 2\sqrt{3}/3}{4}$ $= \frac{\sqrt{16/3}}{2} + \frac{6 - 2\sqrt{3}/3}{4} = \frac{4/\sqrt{3}}{2} + \frac{6 - 2\sqrt{3}/3}{4} = \frac{2}{\sqrt{3}} + \frac{3}{2} - \frac{\sqrt{3}}{6}$ $= \frac{2\sqrt{3}}{3} + \frac{3}{2} - \frac{\sqrt{3}}{6} = \frac{4\sqrt{3} + 9 - \sqrt{3}}{6} = \frac{3\sqrt{3} + 9}{6} = \frac{\sqrt{3} + 3}{2} \approx \frac{1.732+3}{2} = \frac{4.732}{2} = 2.366$ hours. This is indeed the minimum time. 8. **Conclusion:** The man should land the boat approximately $1.155$ miles down the shore from the nearest point to reach the house in the shortest possible time. #### Example Problem 2: Minimizing Cost of Cable A cable company needs to run a cable from a point A on one side of a river to a point B 8 km downstream on the opposite side. The river is 3 km wide. The cost of running cable underwater is $500 per km, and the cost over land is $300 per km. What path should the cable take to minimize the total cost? **Solution:** 1. **Diagram:** A river, points A and B, and a cable path crossing the river then running along the bank. 2. **Variables:** * Let point A be at $(0, 3)$ (assuming river is between $y=0$ and $y=3$). * Let point B be at $(8, 0)$. * Let the cable hit the opposite bank at $(x, 0)$. * River width: 3 km. Distance downstream: 8 km. 3. **Objective Function:** Minimize total cost $C = C_{\text{underwater}} + C_{\text{land}}$. * Distance underwater: $D_{\text{underwater}} = \sqrt{(x-0)^2 + (0-3)^2} = \sqrt{x^2 + 9}$. * Cost underwater: $C_{\text{underwater}} = 500\sqrt{x^2 + 9}$. * Distance over land: $D_{\text{land}} = 8 - x$. * Cost over land: $C_{\text{land}} = 300(8 - x)$. $C(x) = 500\sqrt{x^2 + 9} + 300(8 - x)$. 4. **Domain:** The point $x$ must be between $0$ (crossing directly) and $8$ (crossing to B directly). So, $0 \le x \le 8$. 5. **Differentiate:** $C'(x) = 500 \cdot \frac{1}{2\sqrt{x^2+9}}(2x) + 300(-1)$ $C'(x) = \frac{500x}{\sqrt{x^2+9}} - 300$. 6. **Critical Points:** Set $C'(x) = 0$: $\frac{500x}{\sqrt{x^2+9}} - 300 = 0 \Rightarrow \frac{500x}{\sqrt{x^2+9}} = 300$ $\frac{5x}{\sqrt{x^2+9}} = 3 \Rightarrow 5x = 3\sqrt{x^2+9}$. Square both sides: $(5x)^2 = (3\sqrt{x^2+9})^2 \Rightarrow 25x^2 = 9(x^2+9)$ $25x^2 = 9x^2 + 81 \Rightarrow 16x^2 = 81 \Rightarrow x^2 = \frac{81}{16}$. $x = \sqrt{\frac{81}{16}} = \frac{9}{4} = 2.25$ km (since $x \ge 0$). 7. **Test Critical Points:** $x = 2.25$ is in the domain $[0, 8]$. Using the Second Derivative Test: $C''(x) = 500 \frac{\sqrt{x^2+9}(1) - x \cdot \frac{1}{2\sqrt{x^2+9}}(2x)}{x^2+9} = 500 \frac{(x^2+9) - x^2}{(x^2+9)^{3/2}} = 500 \frac{9}{(x^2+9)^{3/2}} = \frac{4500}{(x^2+9)^{3/2}}$. Since $C''(x) > 0$ for all $x$, the critical point is a local minimum. Also check endpoints: $C(0) = 500\sqrt{0^2+9} + 300(8-0) = 500(3) + 300(8) = 1500 + 2400 = 3900$. $C(8) = 500\sqrt{8^2+9} + 300(8-8) = 500\sqrt{64+9} + 0 = 500\sqrt{73} \approx 500(8.544) \approx 4272$. $C\left(\frac{9}{4}\right) = 500\sqrt{\left(\frac{9}{4}\right)^2+9} + 300\left(8-\frac{9}{4}\right)$ $= 500\sqrt{\frac{81}{16}+\frac{144}{16}} + 300\left(\frac{32-9}{4}\right)$ $= 500\sqrt{\frac{225}{16}} + 300\left(\frac{23}{4}\right) = 500\left(\frac{15}{4}\right) + 75(23)$ $= 125(15) + 1725 = 1875 + 1725 = 3600$. This is the minimum cost. 8. **Conclusion:** The cable should hit the opposite bank $2.25$ km downstream from point A's opposite point to minimize cost. The minimum cost is $3600. #### Example Problem 3: Optimal Production Level A company manufactures $x$ units of a product per day. The cost function is $C(x) = 2000 + 3x + 0.01x^2$. The demand function (price per unit) is $p(x) = 10 - 0.002x$. Find the production level that maximizes profit. **Solution:** 1. **Diagram:** Conceptual graph of cost, revenue, and profit. 2. **Variables:** $x$: number of units produced. 3. **Objective Function:** Maximize Profit $P(x) = R(x) - C(x)$. * Revenue $R(x) = x \cdot p(x) = x(10 - 0.002x) = 10x - 0.002x^2$. * Cost $C(x) = 2000 + 3x + 0.01x^2$. $P(x) = (10x - 0.002x^2) - (2000 + 3x + 0.01x^2)$ $P(x) = 10x - 0.002x^2 - 2000 - 3x - 0.01x^2$ $P(x) = -0.012x^2 + 7x - 2000$. 4. **Domain:** $x \ge 0$. Also, price must be non-negative: $10 - 0.002x \ge 0 \Rightarrow 10 \ge 0.002x \Rightarrow x \le \frac{10}{0.002} = 5000$. So, $0 \le x \le 5000$. 5. **Differentiate:** $P'(x) = -0.024x + 7$. 6. **Critical Points:** Set $P'(x) = 0$: $-0.024x + 7 = 0 \Rightarrow 0.024x = 7 \Rightarrow x = \frac{7}{0.024} = \frac{7000}{24} = \frac{1750}{6} = \frac{875}{3} \approx 291.67$. Since $x$ must be an integer, we might need to check $x=291$ and $x=292$. 7. **Test Critical Points:** $x = \frac{875}{3}$ is in the domain $[0, 5000]$. Using the Second Derivative Test: $P''(x) = -0.024$. Since $P''(x) **Solution:** 1. **Diagram:** A sawtooth graph showing inventory level over time. 2. **Variables:** * $x$: number of units ordered each time. * Annual demand: $D = 1000$ units. * Storage cost per unit per year: $S = 10$. * Ordering cost per order: $O = 50$. 3. **Objective Function:** Minimize total annual inventory cost $TC(x) = \text{Annual Ordering Cost} + \text{Annual Storage Cost}$. * Number of orders per year: $\frac{D}{x} = \frac{1000}{x}$. * Annual ordering cost: $\frac{1000}{x} \cdot O = \frac{1000}{x} \cdot 50 = \frac{50000}{x}$. * Average inventory level (assuming inventory depletes linearly): $\frac{x}{2}$. * Annual storage cost: $\frac{x}{2} \cdot S = \frac{x}{2} \cdot 10 = 5x$. $TC(x) = \frac{50000}{x} + 5x$. 4. **Domain:** $x > 0$. 5. **Differentiate:** $TC'(x) = -\frac{50000}{x^2} + 5$. 6. **Critical Points:** Set $TC'(x) = 0$: $-\frac{50000}{x^2} + 5 = 0 \Rightarrow 5 = \frac{50000}{x^2} \Rightarrow 5x^2 = 50000 \Rightarrow x^2 = 10000 \Rightarrow x = 100$. 7. **Test Critical Points:** $x=100$ is in the domain $x > 0$. Using the Second Derivative Test: $TC''(x) = \frac{100000}{x^3}$. $TC''(100) = \frac{100000}{100^3} = \frac{100000}{1000000} = \frac{1}{10} > 0$. This indicates a local minimum. 8. **Conclusion:** The store should order 100 units each time to minimize total inventory costs. Minimum cost: $TC(100) = \frac{50000}{100} + 5(100) = 500 + 500 = 1000$. #### Example Problem 5: Optimal Speed for Fuel Efficiency The cost of running a truck is $(0.002v^2 + \frac{100}{v})$ dollars per mile, where $v$ is the speed in mph. At what speed should the truck be driven to minimize the cost per mile? **Solution:** 1. **Diagram:** Conceptual graph of cost per mile vs. speed. 2. **Variables:** $v$: speed in mph. 3. **Objective Function:** Minimize Cost per mile $C(v) = 0.002v^2 + \frac{100}{v}$. 4. **Domain:** $v > 0$. 5. **Differentiate:** $C'(v) = 0.004v - \frac{100}{v^2}$. 6. **Critical Points:** Set $C'(v) = 0$: $0.004v - \frac{100}{v^2} = 0 \Rightarrow 0.004v = \frac{100}{v^2} \Rightarrow 0.004v^3 = 100 \Rightarrow v^3 = \frac{100}{0.004} = \frac{100}{4/1000} = \frac{100000}{4} = 25000$. $v = \sqrt[3]{25000} = \sqrt[3]{125 \cdot 200} = 5\sqrt[3]{200} \approx 5(5.848) \approx 29.24$ mph. 7. **Test Critical Points:** $v \approx 29.24$ is in the domain $v > 0$. Using the Second Derivative Test: $C''(v) = 0.004 + \frac{200}{v^3}$. Since $v > 0$, $C''(v)$ is always positive, indicating a local minimum. 8. **Conclusion:** The truck should be driven at approximately $29.24$ mph to minimize the cost per mile. #### Example Problem 6: Maximum Power Transfer The power $P$ delivered to a load resistor $R_L$ by a source with internal resistance $R_S$ is given by $P = \frac{V^2 R_L}{(R_S + R_L)^2}$, where $V$ is the source voltage (constant). What value of $R_L$ maximizes the power delivered? **Solution:** 1. **Diagram:** A simple circuit with source, internal resistance, and load. 2. **Variables:** $R_L$: load resistance. $V, R_S$: constants. 3. **Objective Function:** Maximize $P(R_L) = V^2 R_L (R_S + R_L)^{-2}$. 4. **Domain:** $R_L > 0$. 5. **Differentiate:** Use the product rule. $P'(R_L) = V^2 \left[ 1 \cdot (R_S + R_L)^{-2} + R_L \cdot (-2)(R_S + R_L)^{-3} \cdot (1) \right]$ $P'(R_L) = V^2 (R_S + R_L)^{-3} \left[ (R_S + R_L) - 2R_L \right]$ $P'(R_L) = V^2 (R_S + R_L)^{-3} (R_S - R_L)$. 6. **Critical Points:** Set $P'(R_L) = 0$. Since $V^2 > 0$ and $(R_S + R_L)^{-3} > 0$, we must have: $R_S - R_L = 0 \Rightarrow R_L = R_S$. 7. **Test Critical Points:** If $R_L 0$, so $P'(R_L) > 0$. If $R_L > R_S$: $R_S - R_L **Solution:** 1. **Diagram:** A wall, a painting, a viewer. Use angles. 2. **Variables:** * $x$: distance from the wall (viewer's position). * Height of viewer's eyes: $E = 5$ feet. * Lower edge of painting: $L_1 = 8$ feet from floor. * Upper edge of painting: $L_2 = 8+4 = 12$ feet from floor. * Let $\alpha$ be the angle from the viewer's eyes to the bottom of the painting. * Let $\beta$ be the angle from the viewer's eyes to the top of the painting. * Objective: Maximize viewing angle $\theta = \beta - \alpha$. 3. **Relationships:** Use tangent. * Vertical distance from eyes to lower edge: $8 - 5 = 3$ feet. So, $\tan \alpha = \frac{3}{x}$. * Vertical distance from eyes to upper edge: $12 - 5 = 7$ feet. So, $\tan \beta = \frac{7}{x}$. * $\theta = \arctan\left(\frac{7}{x}\right) - \arctan\left(\frac{3}{x}\right)$. 4. **Domain:** $x > 0$. 5. **Differentiate:** $\theta'(x) = \frac{1}{1 + (7/x)^2} \cdot \left(-\frac{7}{x^2}\right) - \frac{1}{1 + (3/x)^2} \cdot \left(-\frac{3}{x^2}\right)$ $\theta'(x) = \frac{-7/x^2}{(x^2+49)/x^2} + \frac{3/x^2}{(x^2+9)/x^2}$ $\theta'(x) = \frac{-7}{x^2+49} + \frac{3}{x^2+9}$. 6. **Critical Points:** Set $\theta'(x) = 0$: $\frac{3}{x^2+9} = \frac{7}{x^2+49}$ $3(x^2+49) = 7(x^2+9)$ $3x^2 + 147 = 7x^2 + 63$ $84 = 4x^2 \Rightarrow x^2 = 21 \Rightarrow x = \sqrt{21} \approx 4.58$ feet. 7. **Test Critical Points:** If $x 0$. If $x > \sqrt{21}$, e.g., $x=10$: $\theta'(10) = \frac{-7}{149} + \frac{3}{109} \approx -0.046 + 0.027 = -0.019 **Solution:** 1. **Diagram:** Conceptual graph of revenue vs. price. 2. **Variables:** * $p$: ticket price. * $x$: number of attendees. 3. **Relationships:** * When $p=20$, $x=1000$. * For every $1 increase in $p$, $x$ decreases by $50$. We can write $x$ as a linear function of $p$: $x - 1000 = -50(p - 20)$. $x = 1000 - 50p + 1000 = 2000 - 50p$. 4. **Objective Function:** Maximize Revenue $R(p) = p \cdot x = p(2000 - 50p) = 2000p - 50p^2$. 5. **Domain:** $p \ge 0$. Also, number of attendees must be non-negative: $2000 - 50p \ge 0 \Rightarrow 2000 \ge 50p \Rightarrow p \le 40$. So, $0 \le p \le 40$. 6. **Differentiate:** $R'(p) = 2000 - 100p$. 7. **Critical Points:** Set $R'(p) = 0$: $2000 - 100p = 0 \Rightarrow 100p = 2000 \Rightarrow p = 20$. 8. **Test Critical Points:** $p=20$ is in the domain $[0, 40]$. Using the Second Derivative Test: $R''(p) = -100$. Since $R''(p) **Solution:** 1. **Diagram:** An open-top box with a square base. 2. **Variables:** Let $s$ be the side length of the square base, and $h$ be the height. 3. **Objective Function:** Minimize total cost $C = C_{\text{base}} + C_{\text{sides}}$. * Area of base: $s^2$. Cost of base: $3s^2$. * Area of sides: $4sh$. Cost of sides: $2(4sh) = 8sh$. $C(s,h) = 3s^2 + 8sh$. 4. **Constraint:** Volume $V = s^2 h = 10$ cubic meters. 5. **Reduce to Single Variable:** From the constraint, $h = \frac{10}{s^2}$. Substitute $h$ into the cost function: $C(s) = 3s^2 + 8s \left(\frac{10}{s^2}\right) = 3s^2 + \frac{80}{s}$. 6. **Domain:** $s > 0$. 7. **Differentiate:** $C'(s) = 6s - \frac{80}{s^2}$. 8. **Critical Points:** Set $C'(s) = 0$: $6s - \frac{80}{s^2} = 0 \Rightarrow 6s = \frac{80}{s^2} \Rightarrow 6s^3 = 80 \Rightarrow s^3 = \frac{80}{6} = \frac{40}{3}$. $s = \sqrt[3]{\frac{40}{3}} \approx 2.37$ meters. 9. **Test Critical Points:** $s \approx 2.37$ is in the domain $s > 0$. Using the Second Derivative Test: $C''(s) = 6 + \frac{160}{s^3}$. Since $s > 0$, $C''(s)$ is always positive, indicating a local minimum. 10. **Dimensions:** $s = \sqrt[3]{\frac{40}{3}}$ meters $h = \frac{10}{s^2} = \frac{10}{(40/3)^{2/3}} = \frac{10 \cdot 3^{2/3}}{40^{2/3}} = 10 \left(\frac{3}{40}\right)^{2/3} = 10 \left(\frac{120}{1600}\right)^{2/3}$. $h = \frac{10}{(40/3)^{2/3}} = \frac{10}{s^2} = \frac{10}{(40/3) \cdot s^{-1/3}} = \frac{30}{40} s = \frac{3}{4} s = \frac{3}{4} \sqrt[3]{\frac{40}{3}} \approx 1.77$ meters. The dimensions are approximately $2.37 \times 2.37 \times 1.77$ meters. Minimum cost: $C\left(\sqrt[3]{\frac{40}{3}}\right) = 3\left(\frac{40}{3}\right)^{2/3} + 80\left(\frac{40}{3}\right)^{-1/3} = 3\left(\frac{40}{3}\right)^{2/3} + 80 \frac{3^{1/3}}{40^{1/3}}$ $= 3\left(\frac{40}{3}\right)^{2/3} + 2 \cdot 40 \frac{3^{1/3}}{40^{1/3}} = 3\left(\frac{40}{3}\right)^{2/3} + 2 \cdot 40^{2/3} \cdot 3^{1/3}$ This can be simplified: $C(s) = 3s^2 + \frac{80}{s}$. Since $s = \sqrt[3]{40/3}$, $s^3 = 40/3$. $C(s) = 3(40/3)^{2/3} + 80(3/40)^{1/3} = 3(40/3)^{2/3} + 80(3/40)^{1/3}$. This is approximately $C \approx 3(2.37)^2 + 80/2.37 \approx 3(5.6169) + 33.755 \approx 16.85 + 33.75 = 50.60$. #### Example Problem 10: Designing a Window A window is in the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 10 meters, find the dimensions that will allow the maximum amount of light to enter (i.e., maximize the area). **Solution:** 1. **Diagram:** A rectangle with a semicircle on top. 2. **Variables:** * Let $2r$ be the width of the rectangle (and diameter of the semicircle). * Let $h$ be the height of the rectangular part. 3. **Objective Function:** Maximize Area $A = A_{\text{rectangle}} + A_{\text{semicircle}}$. * $A_{\text{rectangle}} = (2r)h$. * $A_{\text{semicircle}} = \frac{1}{2}\pi r^2$. $A(r,h) = 2rh + \frac{1}{2}\pi r^2$. 4. **Constraint:** Perimeter $P = 10$ meters. * Perimeter of rectangle (excluding top side): $2h + 2r$. * Perimeter of semicircle (arc length): $\frac{1}{2}(2\pi r) = \pi r$. $P = 2h + 2r + \pi r = 10$. 5. **Reduce to Single Variable:** From the constraint, $2h = 10 - 2r - \pi r \Rightarrow h = 5 - r - \frac{\pi}{2}r$. Substitute $h$ into the area function: $A(r) = 2r\left(5 - r - \frac{\pi}{2}r\right) + \frac{1}{2}\pi r^2$ $A(r) = 10r - 2r^2 - \pi r^2 + \frac{1}{2}\pi r^2$ $A(r) = 10r - 2r^2 - \frac{1}{2}\pi r^2$. 6. **Domain:** $r > 0$. Also, $h > 0 \Rightarrow 5 - r - \frac{\pi}{2}r > 0 \Rightarrow 5 > r(1 + \frac{\pi}{2}) \Rightarrow r ### Practice Problems (50 problems with solutions) #### Set 1: Basic Max/Min 1. **Problem:** Find two positive numbers whose sum is 20 and whose product is as large as possible. **Solution:** Let the numbers be $x$ and $y$. Constraint: $x+y=20 \Rightarrow y=20-x$. Objective: Maximize $P=xy = x(20-x) = 20x-x^2$. $P'(x)=20-2x=0 \Rightarrow x=10$. $P''(x)=-2 0$). $S''(x)=200/x^3>0$ for $x>0$, so it's a minimum. If $x=10$, $y=10$. Numbers are 10 and 10. 3. **Problem:** A rectangular field is to be enclosed by 400 feet of fencing. What is the maximum possible area? **Solution:** Let $L$ and $W$ be length and width. Perimeter $2L+2W=400 \Rightarrow L+W=200 \Rightarrow W=200-L$. Area $A=LW = L(200-L) = 200L-L^2$. $A'(L)=200-2L=0 \Rightarrow L=100$. $W=100$. Max area $100 \times 100 = 10000$ sq ft. 4. **Problem:** Find the maximum value of $f(x) = x^3 - 6x^2 + 9x + 15$ on the interval $[0, 5]$. **Solution:** $f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$. Critical points at $x=1, x=3$. Evaluate $f(0)=15$, $f(1)=1-6+9+15=19$, $f(3)=27-54+27+15=15$, $f(5)=125-150+45+15=35$. Max value is 35. 5. **Problem:** Find the minimum value of $f(x) = x^4 - 4x^3$ on the interval $[-1, 4]$. **Solution:** $f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$. Critical points at $x=0, x=3$. Evaluate $f(-1)=(-1)^4-4(-1)^3=1+4=5$. $f(0)=0$. $f(3)=3^4-4(3)^3 = 81-108 = -27$. $f(4)=4^4-4(4)^3=256-256=0$. Min value is -27. #### Set 2: Volume/Surface Area 6. **Problem:** A cylindrical can without a top is to have a volume of $100\pi$ cm³. Find the dimensions that will minimize the surface area. **Solution:** $V=\pi r^2 h = 100\pi \Rightarrow h=100/r^2$. Surface Area $A=\pi r^2 + 2\pi rh = \pi r^2 + 2\pi r(100/r^2) = \pi r^2 + 200\pi/r$. $A'(r)=2\pi r - 200\pi/r^2=0 \Rightarrow 2\pi r^3 = 200\pi \Rightarrow r^3=100 \Rightarrow r=\sqrt[3]{100}$. $h=100/(\sqrt[3]{100})^2 = 100/\sqrt[3]{10000} = \sqrt[3]{100}$. So $r=h=\sqrt[3]{100}$ cm. 7. **Problem:** An open-top box is made by cutting squares from the corners of a 20x20 cm cardboard sheet and folding up the sides. What is the maximum volume? **Solution:** Let $x$ be the side of the cut squares. Dimensions of box: $(20-2x) \times (20-2x) \times x$. Volume $V(x) = x(20-2x)^2$. $V'(x) = (20-2x)^2 + x \cdot 2(20-2x)(-2) = (20-2x)(20-2x-4x) = (20-2x)(20-6x)$. Setting $V'(x)=0$ gives $x=10$ or $x=20/6=10/3$. $x=10$ gives zero volume. So $x=10/3$. Max volume $V(10/3) = (10/3)(20-20/3)^2 = (10/3)(40/3)^2 = (10/3)(1600/9) = 16000/27 \approx 592.59$ cm³. 8. **Problem:** Find the dimensions of a right circular cone with slant height 10 cm that has the maximum volume. **Solution:** Let $r$ be radius and $h$ be height. Slant height $L=10$. $r^2+h^2=10^2=100 \Rightarrow r^2=100-h^2$. Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (100-h^2)h = \frac{1}{3}\pi (100h-h^3)$. $V'(h) = \frac{1}{3}\pi (100-3h^2)=0 \Rightarrow 3h^2=100 \Rightarrow h=\sqrt{100/3}=10/\sqrt{3}$. $r^2=100-(10/\sqrt{3})^2=100-100/3=200/3 \Rightarrow r=\sqrt{200/3}=10\sqrt{2}/\sqrt{3}$. 9. **Problem:** A sphere has radius $R$. Find the volume of the largest cone that can be inscribed in the sphere. **Solution:** Let $r$ be the radius of cone base, $h$ be its height. Let $x$ be the distance from sphere center to cone base. $r^2+x^2=R^2$. Cone height $H=R+x$. Volume $V=\frac{1}{3}\pi r^2 H = \frac{1}{3}\pi (R^2-x^2)(R+x) = \frac{1}{3}\pi (R-x)(R+x)^2$. $V'(x) = \frac{1}{3}\pi [-(R+x)^2 + (R-x)2(R+x)] = \frac{1}{3}\pi (R+x)[-(R+x)+2R-2x] = \frac{1}{3}\pi (R+x)[R-3x]$. Setting $V'(x)=0 \Rightarrow x=-R$ (not possible) or $x=R/3$. $H=R+R/3=4R/3$. $r^2=R^2-(R/3)^2=8R^2/9$. Max Volume $V=\frac{1}{3}\pi (8R^2/9)(4R/3) = \frac{32\pi R^3}{81}$. 10. **Problem:** Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius $R$. **Solution:** Let $(x,y)$ be the top-right corner. Width $2x$, height $2y$. Constraint $x^2+y^2=R^2 \Rightarrow y=\sqrt{R^2-x^2}$. Area $A=4xy = 4x\sqrt{R^2-x^2}$. Maximize $A^2 = 16x^2(R^2-x^2) = 16R^2x^2-16x^4$. Derivative wrt $x$: $32R^2x-64x^3=0 \Rightarrow 32x(R^2-2x^2)=0$. $x=R/\sqrt{2}$. $y=\sqrt{R^2-R^2/2}=R/\sqrt{2}$. Dimensions are $2(R/\sqrt{2}) \times 2(R/\sqrt{2})$, which is a square with side $R\sqrt{2}$. #### Set 3: Distance/Geometry 11. **Problem:** Find the point on the line $y=3x+1$ that is closest to the origin. **Solution:** Distance squared $D^2 = x^2+y^2 = x^2+(3x+1)^2 = x^2+9x^2+6x+1 = 10x^2+6x+1$. $(D^2)'=20x+6=0 \Rightarrow x=-6/20=-3/10$. $y=3(-3/10)+1 = -9/10+10/10=1/10$. Point is $(-3/10, 1/10)$. 12. **Problem:** A wire 20 cm long is cut into two pieces. One piece is bent into a square and the other into a circle. How should the wire be cut to minimize the sum of the areas? **Solution:** Let $x$ be length for square, $20-x$ for circle. Side of square $s=x/4$. Radius of circle $2\pi r = 20-x \Rightarrow r=(20-x)/(2\pi)$. Area $A=(x/4)^2 + \pi((20-x)/(2\pi))^2 = x^2/16 + (20-x)^2/(4\pi)$. $A'(x)=2x/16 + 2(20-x)(-1)/(4\pi) = x/8 - (20-x)/(2\pi)=0$. $x/8 = (20-x)/(2\pi) \Rightarrow \pi x = 8(20-x) = 160-8x \Rightarrow (\pi+8)x=160 \Rightarrow x=160/(\pi+8)$. This value will minimize the area. (Check endpoints: $x=0$ (all circle), $x=20$ (all square)). 13. **Problem:** Find the dimensions of the rectangle of largest area that can be inscribed in the region bounded by $y=0$, $x=0$, and $y = 6-x$. **Solution:** The region is a right triangle with vertices $(0,0), (6,0), (0,6)$. Let $(x,y)$ be the top-right corner of the rectangle. Area $A=xy$. Constraint $y=6-x$. $A(x)=x(6-x)=6x-x^2$. $A'(x)=6-2x=0 \Rightarrow x=3$. $y=6-3=3$. Dimensions are $3 \times 3$. Max area 9. 14. **Problem:** Find the dimensions of a rectangle of perimeter $P$ that has the maximum area. **Solution:** $2L+2W=P \Rightarrow W=P/2-L$. $A=L(P/2-L)=PL/2-L^2$. $A'(L)=P/2-2L=0 \Rightarrow L=P/4$. $W=P/2-P/4=P/4$. It's a square with side $P/4$. 15. **Problem:** A rectangular page is to contain 30 square inches of print. The margins at the top and bottom are 1 inch, and the margins at the sides are 0.5 inch. Find the dimensions of the page so that the least amount of paper is used. **Solution:** Let print area dimensions be $x \times y$. So $xy=30 \Rightarrow y=30/x$. Page dimensions are $(x+1) \times (y+2)$. Total Area $A=(x+1)(y+2) = (x+1)(30/x+2) = 30+2x+30/x+2 = 32+2x+30/x$. $A'(x)=2-30/x^2=0 \Rightarrow 2x^2=30 \Rightarrow x^2=15 \Rightarrow x=\sqrt{15}$. $y=30/\sqrt{15} = 2\sqrt{15}$. Page dimensions: $x+1 = \sqrt{15}+1$, $y+2 = 2\sqrt{15}+2$. #### Set 4: Time/Rates/Economics 16. **Problem:** A production function is given by $P(L) = -L^3 + 9L^2 + 12L$, where $L$ is the number of labor units. Find the number of labor units that maximizes production. **Solution:** $P'(L) = -3L^2 + 18L + 12 = -3(L^2-6L-4)$. Roots of $L^2-6L-4=0$ are $L = \frac{6 \pm \sqrt{36-4(-4)}}{2} = \frac{6 \pm \sqrt{52}}{2} = 3 \pm \sqrt{13}$. Since $L>0$, $L=3+\sqrt{13} \approx 6.6$. $P''(L)=-6L+18$. $P''(3+\sqrt{13}) = -6(3+\sqrt{13})+18 = -18-6\sqrt{13}+18 = -6\sqrt{13} 0$, so it's a minimum. $x=2$. 29. **Problem:** A person wants to cross a 3-mile wide river and reach a point 8 miles downstream on the opposite bank. They can row at 4 mph and run at 5 mph. What is the minimum time to reach the destination? **Solution:** This is similar to the boat problem. River width $W=3$. Downstream distance $D=8$. Row speed $v_r=4$. Run speed $v_w=5$. Let $x$ be the distance downstream from the point opposite the starting point where they land. Time $T(x) = \frac{\sqrt{x^2+W^2}}{v_r} + \frac{D-x}{v_w} = \frac{\sqrt{x^2+9}}{4} + \frac{8-x}{5}$. $T'(x) = \frac{x}{4\sqrt{x^2+9}} - \frac{1}{5} = 0 \Rightarrow 5x = 4\sqrt{x^2+9}$. $25x^2 = 16(x^2+9) \Rightarrow 25x^2 = 16x^2+144 \Rightarrow 9x^2=144 \Rightarrow x^2=16 \Rightarrow x=4$. This value is in range $[0,8]$. Minimum time $T(4) = \frac{\sqrt{4^2+9}}{4} + \frac{8-4}{5} = \frac{\sqrt{25}}{4} + \frac{4}{5} = \frac{5}{4} + \frac{4}{5} = \frac{25+16}{20} = \frac{41}{20} = 2.05$ hours. 30. **Problem:** Find the maximum volume of a cylinder that can be inscribed in a sphere of radius $R$. **Solution:** This is Example Problem 4 in Volume Optimization. $h = \frac{2R}{\sqrt{3}}$, $r = R\sqrt{2/3}$. Max volume $V = \frac{4\pi R^3}{3\sqrt{3}}$. #### Set 7: Optimization with Constraints 31. **Problem:** A rectangular box with a square base and no top is to be made from 1200 cm² of material. Find the dimensions that maximize the volume. **Solution:** Base side $s$, height $h$. Surface Area $A = s^2+4sh=1200 \Rightarrow h = (1200-s^2)/(4s)$. Volume $V=s^2 h = s^2 (1200-s^2)/(4s) = (1200s-s^3)/4 = 300s-s^3/4$. $V'(s)=300-3s^2/4=0 \Rightarrow 3s^2=1200 \Rightarrow s^2=400 \Rightarrow s=20$. $h=(1200-400)/(4 \cdot 20) = 800/80=10$. Dimensions are $20 \times 20 \times 10$ cm. 32. **Problem:** A rectangular box with a square base and a top is to have a volume of 1000 cm³. The material for the top and bottom costs $2 per cm², and the material for the sides costs $1 per cm². Find the dimensions that minimize the total cost. **Solution:** Base side $s$, height $h$. Volume $s^2 h = 1000 \Rightarrow h=1000/s^2$. Cost $C = 2(2s^2) + 1(4sh) = 4s^2+4sh = 4s^2+4s(1000/s^2) = 4s^2+4000/s$. $C'(s)=8s-4000/s^2=0 \Rightarrow 8s^3=4000 \Rightarrow s^3=500 \Rightarrow s=\sqrt[3]{500}=5\sqrt[3]{4}$. $h=1000/(5\sqrt[3]{4})^2 = 1000/(25 \cdot 4^{2/3}) = 40/4^{2/3} = 40 \cdot 4^{1/3}/4 = 10 \cdot 4^{1/3} = 10\sqrt[3]{4}$. Dimensions $5\sqrt[3]{4} \times 5\sqrt[3]{4} \times 10\sqrt[3]{4}$ cm. 33. **Problem:** An object is thrown vertically upward with an initial velocity of $v_0$ from an initial height of $s_0$. Its height at time $t$ is $s(t) = -16t^2 + v_0 t + s_0$. If $v_0=64$ ft/s and $s_0=0$, what is the maximum height reached by the object? **Solution:** $s(t) = -16t^2+64t$. $s'(t)=-32t+64=0 \Rightarrow t=2$. $s''(t)=-32 0$ for $x>0$. It's a local minimum. 38. **Problem:** Find the absolute maximum and minimum of $f(x)=x^2-4x+3$ on $[0,4]$. **Solution:** $f'(x)=2x-4=0 \Rightarrow x=2$. $f(0)=3$. $f(2)=4-8+3=-1$. $f(4)=16-16+3=3$. Absolute max is 3, absolute min is -1. 39. **Problem:** The sum of two non-negative numbers is 10. Find the maximum value of the sum of their squares. **Solution:** $x+y=10 \Rightarrow y=10-x$. Sum of squares $S=x^2+y^2=x^2+(10-x)^2 = x^2+100-20x+x^2=2x^2-20x+100$. $S'(x)=4x-20=0 \Rightarrow x=5$. This gives $y=5$, sum of squares $25+25=50$. Consider endpoints: $x=0, y=10 \Rightarrow S=100$. $x=10, y=0 \Rightarrow S=100$. Max value is 100. (The critical point was a local minimum for the sum of squares, not maximum). 40. **Problem:** Find the minimum value of $f(x)=x^2+\frac{2}{x}$ for $x>0$. **Solution:** $f'(x)=2x-\frac{2}{x^2}=0 \Rightarrow 2x^3=2 \Rightarrow x^3=1 \Rightarrow x=1$. $f''(x)=2+\frac{4}{x^3}$. $f''(1)=6>0$, so it's a minimum. Min value $f(1)=1+2=3$. #### Set 9: Applied Problems 41. **Problem:** A company wishes to run a pipeline from point A on one side of a 2 km wide river to point B on the opposite side, 10 km downstream. The cost of laying pipe under water is twice the cost of laying pipe over land. What path should the pipeline take to minimize cost? **Solution:** Let land cost be $k$ per km. Underwater cost is $2k$ per km. River width $W=2$. Downstream $D=10$. Let $x$ be the distance downstream from the point opposite A where pipe comes ashore. Cost $C(x) = 2k\sqrt{x^2+W^2} + k(D-x) = 2k\sqrt{x^2+4} + k(10-x)$. $C'(x) = 2k \frac{x}{\sqrt{x^2+4}} - k = 0 \Rightarrow \frac{2x}{\sqrt{x^2+4}} = 1 \Rightarrow 2x = \sqrt{x^2+4}$. $4x^2 = x^2+4 \Rightarrow 3x^2=4 \Rightarrow x=2/\sqrt{3}$. This is in range $[0,10]$. So, $2/\sqrt{3}$ km downstream. 42. **Problem:** A rectangular piece of land is to be enclosed by a fence. One side of the land is along a straight wall, so no fence is needed there. The area of the land is 1000 m². What is the minimum length of fencing required? **Solution:** Let $L$ be length parallel to wall, $W$ be width. Area $LW=1000 \Rightarrow L=1000/W$. Fencing $F=L+2W = 1000/W+2W$. $F'(W)=-1000/W^2+2=0 \Rightarrow 2W^2=1000 \Rightarrow W^2=500 \Rightarrow W=\sqrt{500}=10\sqrt{5}$. $L=1000/(10\sqrt{5})=100/\sqrt{5}=20\sqrt{5}$. Min fencing $F=20\sqrt{5}+2(10\sqrt{5})=40\sqrt{5} \approx 89.44$ m. 43. **Problem:** Find the maximum volume of a right circular cylinder that can be inscribed in a sphere of radius $R$. **Solution:** This is Example Problem 4 in Volume Optimization. $h = \frac{2R}{\sqrt{3}}$, $r = R\sqrt{2/3}$. Max volume $V = \frac{4\pi R^3}{3\sqrt{3}}$. 44. **Problem:** A conical paper cup is to be made to hold a volume of $V$ cm³. Find the ratio of the height to the radius that minimizes the amount of paper used (i.e., minimizes the surface area). **Solution:** Volume $V=\frac{1}{3}\pi r^2 h$. Surface Area $A=\pi r L = \pi r \sqrt{r^2+h^2}$. From volume, $h=\frac{3V}{\pi r^2}$. $A(r) = \pi r \sqrt{r^2 + \left(\frac{3V}{\pi r^2}\right)^2} = \pi r \sqrt{r^2 + \frac{9V^2}{\pi^2 r^4}}$. Maximize $A^2 = \pi^2 r^2 (r^2 + \frac{9V^2}{\pi^2 r^4}) = \pi^2 r^4 + \frac{9V^2}{r^2}$. Let $f(r) = \pi^2 r^4 + 9V^2 r^{-2}$. $f'(r) = 4\pi^2 r^3 - 18V^2 r^{-3} = 0 \Rightarrow 4\pi^2 r^3 = 18V^2 r^{-3} \Rightarrow 4\pi^2 r^6 = 18V^2 \Rightarrow r^6 = \frac{9V^2}{2\pi^2}$. $h^2 = \frac{9V^2}{\pi^2 r^4} = \frac{9V^2}{\pi^2 (\frac{9V^2}{2\pi^2})^{2/3}} = \frac{9V^2}{\pi^2 \frac{9^{2/3}V^{4/3}}{2^{2/3}\pi^{4/3}}} = \frac{9V^2 \cdot 2^{2/3}\pi^{4/3}}{\pi^2 9^{2/3}V^{4/3}} = \frac{9^{1/3}V^{2/3}2^{2/3}}{\pi^{2/3}}$. This is becoming complex. An easier way: $h^2 = \frac{9V^2}{\pi^2 r^4}$. Substitute $V = \frac{1}{3}\pi r^2 h$: $h^2 = \frac{9(\frac{1}{3}\pi r^2 h)^2}{\pi^2 r^4} = \frac{9 \cdot \frac{1}{9}\pi^2 r^4 h^2}{\pi^2 r^4} = h^2$. This is an identity, not helpful. Let's go back to $4\pi^2 r^6 = 18V^2$. From $h=\frac{3V}{\pi r^2}$, substitute $V=\frac{\pi}{3}r^2 h$: $4\pi^2 r^6 = 18 \left(\frac{\pi}{3}r^2 h\right)^2 = 18 \frac{\pi^2}{9} r^4 h^2 = 2\pi^2 r^4 h^2$. $4\pi^2 r^6 = 2\pi^2 r^4 h^2$. Since $r \ne 0$, $4r^2 = 2h^2 \Rightarrow h^2 = 2r^2 \Rightarrow h = \sqrt{2}r$. Ratio $h/r = \sqrt{2}$. 45. **Problem:** Find the point on the parabola $y=x^2$ that is closest to the point $(0,1)$. **Solution:** Distance squared $D^2 = (x-0)^2 + (y-1)^2 = x^2+(x^2-1)^2 = x^2+x^4-2x^2+1 = x^4-x^2+1$. $(D^2)'=4x^3-2x=2x(2x^2-1)=0$. $x=0$ or $x=\pm 1/\sqrt{2}$. If $x=0$, $y=0$. Point $(0,0)$. $D^2=1$. If $x=\pm 1/\sqrt{2}$, $y=(1/\sqrt{2})^2=1/2$. Points $(\pm 1/\sqrt{2}, 1/2)$. $D^2=(1/2)^2-(1/2)+1 = 1/4-1/2+1=3/4$. Minimum distance is at $(\pm 1/\sqrt{2}, 1/2)$. 46. **Problem:** A cylindrical tank with a hemispherical top is to have a volume of $V$ cubic units. Find the dimensions that minimize the surface area. **Solution:** Let $r$ be the radius of cylinder and hemisphere, $h$ be the height of cylinder. Volume $V = \pi r^2 h + \frac{2}{3}\pi r^3$. Surface Area $A = 2\pi r h + \pi r^2 + 2\pi r^2 = 2\pi r h + 3\pi r^2$. From Volume, $h = \frac{V - \frac{2}{3}\pi r^3}{\pi r^2} = \frac{V}{\pi r^2} - \frac{2}{3}r$. Substitute $h$ into $A$: $A(r) = 2\pi r \left(\frac{V}{\pi r^2} - \frac{2}{3}r\right) + 3\pi r^2 = \frac{2V}{r} - \frac{4}{3}\pi r^2 + 3\pi r^2 = \frac{2V}{r} + \frac{5}{3}\pi r^2$. $A'(r) = -\frac{2V}{r^2} + \frac{10}{3}\pi r = 0 \Rightarrow \frac{10}{3}\pi r = \frac{2V}{r^2} \Rightarrow 10\pi r^3 = 6V \Rightarrow r^3 = \frac{3V}{5\pi}$. $r = \sqrt[3]{\frac{3V}{5\pi}}$. $h = \frac{V}{\pi r^2} - \frac{2}{3}r$. Substitute $V = \frac{5}{3}\pi r^3$: $h = \frac{\frac{5}{3}\pi r^3}{\pi r^2} - \frac{2}{3}r = \frac{5}{3}r - \frac{2}{3}r = r$. So, $h=r$. 47. **Problem:** The sum of two positive numbers is 16. What is the minimum possible value of the sum of their cubes? **Solution:** $x+y=16 \Rightarrow y=16-x$. Sum of cubes $S=x^3+y^3 = x^3+(16-x)^3$. $S'(x)=3x^2+3(16-x)^2(-1) = 3x^2-3(16-x)^2 = 3[x^2-(16-x)^2] = 3[x-(16-x)][x+(16-x)] = 3[2x-16][16]$. $S'(x)=48(2x-16)=0 \Rightarrow 2x=16 \Rightarrow x=8$. $y=8$. $S''(x)=48(2)=96>0$, so it's a minimum. Min sum of cubes is $8^3+8^3=512+512=1024$. 48. **Problem:** A farmer has 100 feet of fencing and wants to enclose a rectangular area and divide it into two pens with a fence parallel to one side. What is the maximum area the farmer can enclose? **Solution:** Let $L$ be the length of the outer rectangle, and $W$ be the width. The dividing fence is also of length $W$. Total fencing $2L+3W=100$. Area $A=LW$. From fencing, $L=(100-3W)/2$. $A(W)=W(100-3W)/2 = 50W - \frac{3}{2}W^2$. $A'(W)=50-3W=0 \Rightarrow W=50/3$. $L=(100-3(50/3))/2 = (100-50)/2 = 50/2 = 25$. Max area $A=(25)(50/3) = 1250/3 \approx 416.67$ sq ft. 49. **Problem:** Find the maximum volume of a box with a square base and no top, having a total surface area of $A_0$. **Solution:** Let $s$ be side of base, $h$ be height. Surface Area $s^2+4sh=A_0 \Rightarrow h=(A_0-s^2)/(4s)$. Volume $V=s^2 h = s^2(A_0-s^2)/(4s) = (A_0s-s^3)/4$. $V'(s)=(A_0-3s^2)/4=0 \Rightarrow 3s^2=A_0 \Rightarrow s=\sqrt{A_0/3}$. $h=(A_0-A_0/3)/(4\sqrt{A_0/3}) = (2A_0/3)/(4\sqrt{A_0/3}) = (A_0/6)/\sqrt{A_0/3} = \frac{A_0}{6} \frac{\sqrt{3}}{\sqrt{A_0}} = \frac{\sqrt{A_0}\sqrt{3}}{6} = \frac{\sqrt{3A_0}}{6}$. Max volume $V = \frac{1}{4} (A_0\sqrt{A_0/3} - (A_0/3)\sqrt{A_0/3}) = \frac{1}{4} (\frac{2A_0}{3}\sqrt{A_0/3}) = \frac{A_0}{6}\sqrt{A_0/3} = \frac{A_0\sqrt{A_0}}{6\sqrt{3}} = \frac{A_0^{3/2}\sqrt{3}}{18}$. 50. **Problem:** An airline offers a flight from city A to city B. They charge $300 per ticket and sell 200 tickets. For every $10 decrease in ticket price, they sell 20 more tickets. What ticket price maximizes revenue? **Solution:** Let $p$ be the ticket price. Let $x$ be the number of tickets sold. Initial: $p_0=300, x_0=200$. Change: $-10$ in price, $+20$ in tickets. So for every $1 decrease in price, $20/10=2$ more tickets. Let $k$ be the number of $10 price decreases. Price: $300-10k$. Tickets: $200+20k$. Revenue $R(k) = (300-10k)(200+20k) = 60000 + 6000k - 2000k - 200k^2 = -200k^2 + 4000k + 60000$. $R'(k) = -400k + 4000 = 0 \Rightarrow k=10$. Price $= 300 - 10(10) = 300-100 = 200$. Max revenue $R(10) = (200)(200+200) = 200 \times 400 = 80000$. Alternatively, let $p$ be price. Tickets sold $x = 200 + \frac{20}{10}(300-p) = 200+2(300-p) = 200+600-2p = 800-2p$. Revenue $R(p) = p(800-2p) = 800p-2p^2$. $R'(p) = 800-4p=0 \Rightarrow p=200$. Max revenue at $p=200. ### Important Information for Your Exam Before your exam, ensure you have a solid understanding of these foundational calculus concepts, as they are crucial for solving optimization problems. #### 1. Derivatives * **Power Rule:** $\frac{d}{dx}(x^n) = nx^{n-1}$ * **Product Rule:** $\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$ * **Quotient Rule:** $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$ * **Chain Rule:** $\frac{d}{dx}(f(g(x))) = f'(g(x))g'(x)$ * **Derivatives of Trigonometric Functions:** * $\frac{d}{dx}(\sin x) = \cos x$ * $\frac{d}{dx}(\cos x) = -\sin x$ * $\frac{d}{dx}(\tan x) = \sec^2 x$ * $\frac{d}{dx}(\cot x) = -\csc^2 x$ * $\frac{d}{dx}(\sec x) = \sec x \tan x$ * $\frac{d}{dx}(\csc x) = -\csc x \cot x$ * **Derivatives of Inverse Trigonometric Functions:** * $\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}$ * $\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}$ * **Derivatives of Exponential and Logarithmic Functions:** * $\frac{d}{dx}(e^x) = e^x$ * $\frac{d}{dx}(a^x) = a^x \ln a$ * $\frac{d}{dx}(\ln x) = \frac{1}{x}$ * $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$ #### 2. Critical Points & Extrema * **Critical Point:** A point $c$ in the domain of $f$ where $f'(c) = 0$ or $f'(c)$ is undefined. These are candidates for local maxima or minima. * **First Derivative Test:** * If $f'(x)$ changes from positive to negative at $c$, then $f(c)$ is a local maximum. * If $f'(x)$ changes from negative to positive at $c$, then $f(c)$ is a local minimum. * If $f'(x)$ does not change sign at $c$, then $f(c)$ is neither. * **Second Derivative Test:** * If $f'(c) = 0$ and $f''(c) 0$, then $f(c)$ is a local minimum. * If $f'(c) = 0$ and $f''(c) = 0$, the test is inconclusive. * **Absolute Extrema on a Closed Interval:** To find the absolute maximum and minimum of a continuous function $f$ on a closed interval $[a, b]$: 1. Find the critical points of $f$ in $(a, b)$. 2. Evaluate $f$ at the critical points and at the endpoints $a$ and $b$. 3. The largest of these values is the absolute maximum, and the smallest is the absolute minimum. #### 3. Problem Solving Strategy Revisit the "Steps to Solve Optimization Problems" in the Introduction. Practice applying each step systematically. #### 4. Common Formulas * **Area of a rectangle:** $LW$ * **Perimeter of a rectangle:** $2L+2W$ * **Area of a circle:** $\pi r^2$ * **Circumference of a circle:** $2\pi r$ * **Volume of a box:** $LWH$ * **Volume of a cylinder:** $\pi r^2 h$ * **Surface area of a cylinder (closed):** $2\pi r^2 + 2\pi rh$ * **Volume of a cone:** $\frac{1}{3}\pi r^2 h$ * **Surface area of a cone (open base):** $\pi r L$ (where $L$ is slant height) * **Volume of a sphere:** $\frac{4}{3}\pi r^3$ * **Surface area of a sphere:** $4\pi r^2$ * **Distance formula:** $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ * **Pythagorean theorem:** $a^2+b^2=c^2$ * **Similar triangles:** Ratios of corresponding sides are equal. #### 5. Units Always include appropriate units in your final answers. #### 6. Domain Restrictions Pay close attention to the domain of your objective function. Physical quantities (like length, volume, time) must be positive. This can restrict your search for extrema to a specific interval, requiring you to check endpoints. #### 7. Interpretation After finding a critical point, ensure you interpret it correctly in the context of the problem (e.g., if it's a maximum or minimum, and what quantities it represents). Good luck with your exam!