Intro to Optimization - Calculus 1

1. A person in a boat is 3 km from the nearest point on a straight shoreline. They wish to reach a house 8 km down the shore from that point. They can row at 6 km/h and run along the shore at 10 km/h. To minimize the travel time, to what point on the shore should they row? Let $x$ be the distance in km from the point on the shore nearest the boat to the point where they land.
   Answer: $x = 2.25$ km This is a classic optimization problem where you need to minimize travel time by finding the optimal combination of two different speeds. When you see problems involving "shortest time" or "minimum cost" with multiple rates, think calculus optimization. Let's set up the problem systematically. If the person rows to a point $x$ km along the shore, they travel $\sqrt{x^2 + 9}$ km by boat (using the Pythagorean theorem), then run the remaining $(8-x)$ km along shore. The total time function is: $T(x) = \frac{\sqrt{x^2 + 9}}{6} + \frac{8-x}{10}$ To minimize time, take the derivative and set it equal to zero: $T'(x) = \frac{x}{6\sqrt{x^2 + 9}} - \frac{1}{10} = 0$ Solving: $\frac{x}{6\sqrt{x^2 + 9}} = \frac{1}{10}$ Cross-multiply: $10x = 6\sqrt{x^2 + 9}$ Square both sides: $100x^2 = 36(x^2 + 9)$ This gives us $64x^2 = 324$, so $x^2 = \frac{324}{64} = \frac{81}{16}$ Therefore $x = \frac{9}{4} = 2.25$ km, which is answer C. Answer A (4.00 km) likely comes from incorrectly setting up the distance relationship. Answer B (3.75 km) might result from algebraic errors when solving the optimization equation. Answer D (5.00 km) could come from misunderstanding the constraint that $x$ must be less than 8. Strategy tip: In optimization problems, always verify your derivative algebra carefully—small errors in solving $f'(x) = 0$ lead to wrong answers that often appear as distractors.
2. A cylindrical can with a top and bottom must hold a volume of $1000$ cm$^3$. The material for the top and bottom costs three times as much per unit area as the material for the side. What is the ratio of the height, $h$, to the radius, $r$, that minimizes the cost of the can?
   Answer: $h/r = 6$ Let the cost per unit area for the side be $k$. Then the cost per unit area for the top and bottom is $3k$. The surface area of the side is $2\pi r h$ and the area of the top and bottom combined is $2\pi r^2$. The cost function is $C = k(2\pi r h) + 3k(2\pi r^2) = 2\pi k (rh + 3r^2)$. The volume constraint is $V = \pi r^2 h = 1000$, so $h = 1000/(\pi r^2)$. Substituting into the cost function: $C(r) = 2\pi k (r(1000/(\pi r^2)) + 3r^2) = 2\pi k (1000/(\pi r) + 3r^2)$. To find the minimum, we differentiate with respect to $r$: $C'(r) = 2\pi k (-1000/(\pi r^2) + 6r)$. Setting $C'(r) = 0$ gives $6r = 1000/(\pi r^2)$, which simplifies to $6\pi r^3 = 1000$. Substitute $1000 = \pi r^2 h$: $6\pi r^3 = \pi r^2 h$. Dividing by $\pi r^2$ (since $r\neq0$) gives $6r = h$, so the ratio is $h/r = 6$.
3. The selling price $p$ (in dollars) of an item is related to the number of units sold $x$ by the demand equation $p = 150 - 0.5x$. The cost of producing $x$ units is $C(x) = 4000 + 0.25x^2$. How many units should be sold to maximize profit?
   Answer: 100 The revenue function is $R(x) = xp = x(150 - 0.5x) = 150x - 0.5x^2$. The profit function is $P(x) = R(x) - C(x) = (150x - 0.5x^2) - (4000 + 0.25x^2) = -0.75x^2 + 150x - 4000$. To maximize profit, we find the derivative $P'(x)$ and set it to zero. $P'(x) = -1.5x + 150$. Setting $P'(x) = 0$ gives $1.5x = 150$, so $x = 100$. Since $P''(x) = -1.5 < 0$, this production level yields a maximum profit. A: Correct. B: This is the y-intercept of the demand equation, the price if zero units are sold. C: This might be found by setting revenue equal to cost, a break-even analysis, not a profit maximization. $150x - 0.5x^2 = 4000 + 0.25x^2$ -> $0.75x^2 - 150x + 4000 = 0$. This is hard to solve. Or maybe by setting marginal revenue equal to average cost. A more likely error is to set revenue's derivative to zero: $R'(x) = 150-x=0$ gives $x=150$. This maximizes revenue, not profit. D: This is the x-intercept of the demand equation, the number of units sold if the price is zero.
4. The illumination from a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. Two light sources, one of which is 8 times as strong as the other, are placed 12 feet apart. At what distance from the stronger light source is the total illumination at a minimum?
   Answer: 8 feet Let the strength of the weaker source be $S$ and the stronger source be $8S$. Let a point be at a distance $x$ from the stronger source. Then its distance from the weaker source is $12-x$. The total illumination $I$ at this point is the sum of the illuminations from both sources: $I(x) = \frac{k(8S)}{x^2} + \frac{kS}{(12-x)^2}$, where $k$ is a proportionality constant. We can minimize $f(x) = I(x)/(kS) = \frac{8}{x^2} + \frac{1}{(12-x)^2}$ for $x \in (0, 12)$. We differentiate: $f'(x) = -\frac{16}{x^3} - \frac{-2}{(12-x)^3} = -\frac{16}{x^3} + \frac{2}{(12-x)^3}$. Setting $f'(x) = 0$ gives $\frac{16}{x^3} = \frac{2}{(12-x)^3}$. This simplifies to $8(12-x)^3 = x^3$. Taking the cube root of both sides gives $2(12-x) = x$. So, $24 - 2x = x$, which gives $3x = 24$, and $x = 8$. The minimum illumination occurs at 8 feet from the stronger light source. A: This would mean $x=4$ and $12-x=8$. This results if the cube root is not taken correctly, leading to $2(12-x)=x$ -> $24-2x=x$ -> $x=8$. Wait. This works. What if they solve $8(12-x)=x$? $96-8x=x$ -> $9x=96$. No. What if they take square root instead of cube root? $\sqrt{8}(12-x)=x$ -> $2\sqrt{2}(12-x)=x$. This is too messy. A distance of 4 feet implies $x=4$. Then $12-x=8$. Then $16/4^3 = 16/64 = 1/4$ and $2/8^3 = 2/512 = 1/256$. Not equal. So 4 feet is incorrect. B: 6 feet is halfway, which would be the minimum if the sources had equal strength. D: 9 feet would put the point 3 feet from the weaker source. This might be a ratio error.
5. An open-topped cylindrical container is to be made from $300\pi$ square inches of material. What is the relationship between the height $h$ and the radius $r$ of the container with the largest possible volume?
   Answer: $h = r$ This is a classic optimization problem where you need to maximize volume subject to a constraint on surface area. When you see "largest possible" or "minimum cost" with geometric constraints, you're looking at an optimization problem that requires calculus. To solve this, start by setting up your equations. The surface area constraint is $300\pi = \pi r^2 + 2\pi rh$ (bottom plus curved sides), which gives you $h = \frac{300 - r^2}{2r}$. The volume to maximize is $V = \pi r^2 h = \pi r^2 \cdot \frac{300 - r^2}{2r} = \frac{\pi r(300 - r^2)}{2}$. Taking the derivative and setting it equal to zero: $\frac{dV}{dr} = \frac{\pi(300 - 3r^2)}{2} = 0$. This gives $r^2 = 100$, so $r = 10$. Substituting back: $h = \frac{300 - 100}{20} = 10$. Therefore, $h = r$. Looking at the wrong answers: Choice A ($h = \pi r$) would give an irrational relationship that doesn't emerge from this optimization. Choice B ($h = 2r$) is a common trap - it's the relationship for maximizing surface area given fixed volume, not the reverse. Choice C ($h = r/2$) represents the opposite error, perhaps from confusing which variable should be larger. Study tip: In optimization problems, always clearly identify what you're maximizing/minimizing versus what's constrained. Set up both equations before taking derivatives, and remember that geometric optimization often yields surprisingly simple integer or rational relationships.
6. At $t=0$, a ship is 50 km north of a lighthouse and is sailing south at 20 km/h. At the same time, another ship is at the lighthouse and sailing east at 15 km/h. At what time $t>0$ are the two ships closest to each other?
   Answer: $1.6$ hours Let the lighthouse be at the origin $(0,0)$. At time $t$, the first ship's position is $(0, 50-20t)$. The second ship's position is $(15t, 0)$. The square of the distance between them is $S(t) = (15t - 0)^2 + (0 - (50-20t))^2 = 225t^2 + (50-20t)^2 = 225t^2 + 2500 - 2000t + 400t^2 = 625t^2 - 2000t + 2500$. To minimize the distance, we minimize its square. We find the derivative: $S'(t) = 1250t - 2000$. Setting $S'(t) = 0$ gives $1250t = 2000$, so $t = 2000/1250 = 200/125 = 8/5 = 1.6$ hours. Since $S''(t) = 1250 > 0$, this time corresponds to a minimum distance.
7. A rectangular poster is to have a total area of 180 square inches. The poster consists of a central printed area with a margin of 1 inch on each side and 1.5 inches on the top and bottom. What are the outer dimensions (width and height) of the poster that will maximize the printed area?
   Answer: Width = $2\sqrt{30}$ in, Height = $3\sqrt{30}$ in This is a classic optimization problem where you need to set up variables, express constraints, and use calculus to find maximum values. When you see "maximize" or "minimize" with given constraints, think about setting up a function to optimize. Let's define the outer dimensions as width $w$ and height $h$. Since the total area is 180 square inches, we have $wh = 180$, so $h = \frac{180}{w}$. The printed area is what remains after subtracting the margins. With 1-inch margins on each side, the printed width is $(w-2)$. With 1.5-inch margins on top and bottom, the printed height is $(h-3)$. So the printed area is $A = (w-2)(h-3)$. Substituting our constraint: $A = (w-2)\left(\frac{180}{w}-3\right) = (w-2)\frac{180-3w}{w} = \frac{(w-2)(180-3w)}{w}$ To maximize, take the derivative and set it equal to zero. Using the quotient rule: $A'(w) = \frac{w(-6w+186) - (w-2)(180-3w)}{w^2} = \frac{-3w^2 + 12w + 360}{w^2}$ Setting $A'(w) = 0$: $-3w^2 + 12w + 360 = 0$, which gives $w^2 - 4w - 120 = 0$. Using the quadratic formula: $w = \frac{4 \pm \sqrt{16 + 480}}{2} = \frac{4 \pm \sqrt{496}}{2} = 2 \pm \sqrt{124} = 2 \pm 2\sqrt{31}$. Since $w > 0$, we get $w = 2 + 2\sqrt{30} = 2\sqrt{30}$ and $h = 3\sqrt{30}$. Choice A gives dimensions that are 2 inches less than the correct values. Choice B uses round numbers that don't satisfy the optimization conditions. Choice D also uses convenient numbers but yields suboptimal printed area. When solving optimization problems, always verify your constraint equation and remember that the derivative test finds critical points—don't forget to check that you're truly maximizing.
8. To find the absolute maximum value of a continuous function $f(x)$ on a closed interval $[a, b]$, which of the following procedures is always sufficient?
   Answer: Evaluate $f(x)$ at $x=a$, $x=b$, and all points in $(a,b)$ where $f'(x)=0$ or $f'(x)$ is undefined. The largest of these values is the absolute maximum. The Extreme Value Theorem guarantees that a continuous function on a closed interval has an absolute maximum and minimum. These extrema can occur either at the endpoints of the interval or at critical points within the interval. Critical points are defined as points where the derivative is zero or undefined. Therefore, to be certain of finding the absolute maximum, one must find all critical points, evaluate the function at these points and at the endpoints, and compare all these values. Option A is insufficient as it misses endpoints and points where the derivative is undefined. Option B only identifies local maximums and is also incomplete. Option C is insufficient as it ignores the behavior of the function inside the interval.
9. A particle moves along the x-axis with velocity function $v(t) = t^3 - 6t^2 + 9t + 1$ for $t \ge 0$. What is the maximum acceleration of the particle on the time interval $t \in [0, 4]$?
   Answer: 9 The acceleration function $a(t)$ is the derivative of the velocity function $v(t)$. So, $a(t) = v'(t) = 3t^2 - 12t + 9$. To find the maximum acceleration on the closed interval $[0, 4]$, we need to find the critical points of $a(t)$ and compare the values of $a(t)$ at the critical points and the endpoints. The derivative of acceleration is $a'(t) = 6t - 12$. Setting $a'(t) = 0$ gives $6t - 12 = 0$, so $t = 2$. Now we evaluate $a(t)$ at the critical point $t=2$ and the endpoints $t=0$ and $t=4$. $a(0) = 9$, $a(2) = 12 - 24 + 9 = -3$, $a(4) = 48 - 48 + 9 = 9$. The maximum value is 9. A: This is the minimum velocity on the interval, $v(3) = 1$. B: This is the maximum velocity on the interval. A student might optimize velocity instead of acceleration. C: Correct. D: This is the magnitude of the jerk (rate of change of acceleration) at the endpoints: $|a'(0)| = |a'(4)| = 12$.
10. If $xy + 2x + y = 19$, what is the maximum value of $x$?
    Answer: There is no maximum value. This question looks like an optimization problem but tests the understanding of the underlying function. Let's solve the equation for $y$ in terms of $x$ to see the relationship: $y(x+1) = 19 - 2x$, so $y = \frac{19-2x}{x+1}$. This is a rational function. The question asks for the maximum value of $x$. The domain of this function is all real numbers except $x=-1$. There are no constraints in the problem that would limit the possible values of $x$. For any real number $x \neq -1$, we can find a corresponding $y$ that satisfies the equation. Therefore, $x$ can be arbitrarily large (or arbitrarily small). There is no maximum value for $x$. This is a trick question to make students apply optimization techniques where they are not applicable. A: Correct. The variable $x$ is not bounded. B: This is the constant in the equation and a potential guess. C: This might be found if there was another constraint, or by misinterpreting the question. D: While there is no maximum value, the reason is not that the information is insufficient, but that the variable is unbounded.
11. A rectangular box with a square base and an open top is to have a volume of 32,000 cm³. Find the minimum surface area of the box.
    Answer: 4800 cm² Let the side length of the square base be $x$ and the height be $h$. The volume is $V = x^2 h = 32000$, so $h = 32000/x^2$. The surface area of the open-top box consists of the base and four sides: $A = x^2 + 4xh$. Substituting for $h$, we get $A(x) = x^2 + 4x(32000/x^2) = x^2 + 128000/x$. To minimize the area, we take the derivative: $A'(x) = 2x - 128000/x^2$. Setting $A'(x) = 0$ gives $2x = 128000/x^2$, so $2x^3 = 128000$, which means $x^3 = 64000$. Taking the cube root, we get $x = 40$ cm. The corresponding height is $h = 32000/(40^2) = 32000/1600 = 20$ cm. The minimum surface area is $A(40) = 40^2 + 128000/40 = 1600 + 3200 = 4800$ cm². A: This could be the area of the four sides $4xh = 4(40)(20) = 3200$, or the base area $x^2=1600$. This is likely an arithmetic error. B: This might result from forgetting the base area and making an error, or from using an incorrect formula for volume or area. C: This is the correct minimum surface area. D: This value may come from an algebraic mistake, such as setting $x=20$ and $h=80$, which gives a surface area of $20^2+4(20)(80) = 400+6400=6800$. A value of 6000 is a plausible miscalculation.
12. A rectangular beam is to be cut from a circular log of diameter 20 inches. The strength of the beam is proportional to the product of its width and the square of its depth ($S = kwd^2$). What is the width of the strongest possible beam that can be cut from the log?
    Answer: $\frac{20}{\sqrt{3}}$ inches Let the width of the beam be $w$ and the depth be $d$. A cross-section of the log is a circle with diameter 20, so radius 10. By the Pythagorean theorem, $w^2 + d^2 = (20)^2 = 400$. Thus, $d^2 = 400 - w^2$. The strength is $S(w) = k w d^2 = k w (400 - w^2) = k(400w - w^3)$, where $k$ is a positive constant. The domain for $w$ is $(0, 20)$. To maximize strength, we differentiate: $S'(w) = k(400 - 3w^2)$. Setting $S'(w) = 0$ gives $400 - 3w^2 = 0$, so $3w^2 = 400$, and $w^2 = 400/3$. Since width must be positive, $w = \sqrt{400/3} = 20/\sqrt{3}$. This is the width that maximizes the strength. A: This is the radius of the log, not the optimal width. This might be chosen if $w=d$ is assumed. B: This is the depth $d$ that corresponds to the optimal width: $d^2 = 400 - 400/3 = 800/3 \implies d = \sqrt{800/3} = 20\sqrt{2}/\sqrt{3}$. This is not $10\sqrt{3}$. Let's recheck. $w=10\sqrt{3}$, $w^2=300$. $d^2=100$, $d=10$. Then $S = k(10\sqrt{3})(100) = 1000\sqrt{3} k$. With $w=20/\sqrt{3}$, $d^2=800/3$. $S = k(20/\sqrt{3})(800/3) = 16000k/(3\sqrt{3}) = 16000\sqrt{3}k/9$. Let's compare $1000\sqrt{3}$ with $16000\sqrt{3}/9$. $16000/9 \approx 1777 > 1000$. So my answer is correct. The distractor $10\sqrt{3}$ is the width that would be found if the strength formula was $S=kw^2d$ instead of $S=kwd^2$. D: This is half the correct width, a possible calculation error.
13. A trough is to be made from a long rectangular piece of sheet metal 30 cm wide by bending up one-third of the metal on each side through an angle $\theta$. To maximize the cross-sectional area of the trough, what must be the value of $\cos(\theta)$?
    Answer: 1/2 The sheet metal is 30 cm wide, and is bent up into three sections of 10 cm each. The cross-section is a trapezoid. The base is 10 cm. The two slanted sides are 10 cm. Let $\theta$ be the angle with the vertical. The question says angle $\theta$ but doesn't specify with what. Let's assume it's the angle the sides make with the horizontal. The height of the trapezoid is $h = 10\sin(\theta)$. The top base is $10 + 2(10\cos(\theta))$. The area is $A(\theta) = \frac{1}{2}(base_1 + base_2)h = \frac{1}{2}(10 + 10 + 20\cos(\theta))(10\sin(\theta)) = (10 + 10\cos(\theta))(10\sin(\theta)) = 100(\sin(\theta) + \sin(\theta)\cos(\theta))$. We want to maximize this for $\theta \in [0, \pi/2]$. $A'(\theta) = 100(\cos(\theta) + \cos^2(\theta) - \sin^2(\theta))$. Using $\sin^2(\theta)=1-\cos^2(\theta)$, we get $A'(\theta) = 100(\cos(\theta) + 2\cos^2(\theta) - 1)$. Let $u=\cos(\theta)$. We need to solve $2u^2+u-1=0$, which factors as $(2u-1)(u+1)=0$. So $u = 1/2$ or $u = -1$. Since $\theta \in [0, \pi/2]$, $u=\cos(\theta)$ must be non-negative. Thus, $\cos(\theta) = 1/2$. This corresponds to $\theta = \pi/3$ or 60 degrees. A: Correct. B: $\cos(\theta)=1$ means $\theta=0$, which gives zero area (a flat sheet). C: $\cos(\theta)=\sqrt{3}/2$ corresponds to $\theta=30$ degrees. This is a plausible guess but not optimal. D: $\cos(\theta)=0$ means $\theta=90$ degrees, which forms a rectangle with area $10 \times 10 = 100$. The area for $\theta=60$ is $100(\sin(60)+\sin(60)\cos(60)) = 100(\sqrt{3}/2 + (\sqrt{3}/2)(1/2)) = 100(3\sqrt{3}/4) = 75\sqrt{3} \approx 130 > 100$. So the rectangle is not optimal.
14. Let $f$ be a function such that $f'(x) = (x-1)(x-4)^2$. To find the local extrema of $f$, a student calculates $f''(x) = (x-4)^2 + 2(x-1)(x-4) = (x-4)(3x-6)$. The student notes that $f''(1) = 0$ and concludes the Second Derivative Test is inconclusive at $x=1$. Which of the following is true about $f$ at $x=1$?
    Answer: $f$ has a local minimum at $x=1$. The student correctly found that the Second Derivative Test is inconclusive at $x=1$ because $f''(1)=0$. In this case, we must use the First Derivative Test. The critical points are where $f'(x) = (x-1)(x-4)^2 = 0$, which are $x=1$ and $x=4$. We need to check the sign of $f'(x)$ around $x=1$. The term $(x-4)^2$ is always non-negative. For $x < 1$, the term $(x-1)$ is negative, so $f'(x)$ is negative. For $x$ slightly greater than 1 (but less than 4), the term $(x-1)$ is positive, so $f'(x)$ is positive. Since $f'(x)$ changes sign from negative to positive at $x=1$, the function $f$ has a local minimum at $x=1$. A: This would be true if the sign of $f'(x)$ changed from positive to negative. C: This would be true if the sign of $f'(x)$ did not change at $x=1$ (e.g., if the factor was $(x-1)^2$). D: While the Second Derivative Test is inconclusive, the First Derivative Test is not. This statement is a common misconception that if one test fails, no conclusion can be drawn.
15. A function $g(x)$ is continuous on the interval $[0, 10]$ and has a single critical point at $x=4$. Which of the following pieces of information is sufficient to conclude that $g(x)$ has an absolute minimum at $x=4$?
    Answer: $g(4) < g(0)$ and $g(4) < g(10)$ To find the absolute minimum on a closed interval, we must check the function's values at the critical points and the endpoints. The candidates for the location of the absolute minimum are $x=0$, $x=4$, and $x=10$. B: The statement $g(4) < g(0)$ and $g(4) < g(10)$ directly compares the value of the function at the only critical point with the values at both endpoints. Since $g(4)$ is smaller than the values at the endpoints, and there are no other critical points where a minimum could occur, $g(4)$ must be the absolute minimum value. This is a sufficient condition. A: The condition $g''(4) = 5 > 0$ is the Second Derivative Test, which only implies that $g(x)$ has a local minimum at $x=4$. It is possible that the value of the function at one of the endpoints, $g(0)$ or $g(10)$, is less than $g(4)$. Therefore, this is not sufficient. C: The condition that $g'(x) < 0$ for $x$ in $(0, 4)$ means the function is decreasing as it approaches $x=4$ from the left. This does not tell us anything about the behavior of the function for $x > 4$. The function could continue to decrease, in which case $x=4$ would not be a minimum. This is not sufficient. D: Comparing the values at the endpoints does not provide enough information about the value at the critical point $x=4$. This is not sufficient.
16. A cylindrical can with a closed top and bottom is to be constructed to hold a volume of $16\pi$ cubic centimeters. The material for the top and bottom costs $2 per square centimeter, and the material for the side costs $1 per square centimeter. What is the ratio of height to radius, $h/r$, for the can with the minimum cost?
    Answer: 4 The volume is $V = \pi r^2 h = 16\pi$, so $h = 16/r^2$. The cost function $C$ is given by the cost of the top and bottom plus the cost of the side: $C = 2(2\pi r^2) + 1(2\pi rh) = 4\pi r^2 + 2\pi rh$. Substituting for $h$, we get $C(r) = 4\pi r^2 + 2\pi r(16/r^2) = 4\pi r^2 + 32\pi/r$. To minimize cost, we find the derivative: $C'(r) = 8\pi r - 32\pi/r^2$. Setting $C'(r) = 0$ gives $8\pi r = 32\pi/r^2$, which simplifies to $r^3 = 4$, so $r = \sqrt[3]{4}$. The corresponding height is $h = 16/r^2 = 16/(\sqrt[3]{4})^2 = 16/4^{2/3} = 4^2/4^{2/3} = 4^{4/3}$. The ratio is $h/r = 4^{4/3} / 4^{1/3} = 4^{(4/3 - 1/3)} = 4^1 = 4$. A: This ratio would be found if the costs were equal and the quantity to be optimized was $r+h$ or something similar, likely an algebraic error. B: This is the well-known ratio that minimizes the surface area of a cylinder for a fixed volume, a common trap for students who have memorized that result without considering the differential costs. D: This ratio may result from an algebraic error, such as setting $r^3=8$ instead of $r^3=4$.
17. A company's profit $P(x)$ from selling $x$ units of a product is given by $P(x) = R(x) - C(x)$, where $R(x)$ is revenue and $C(x)$ is cost. The demand function is $p(x) = 500 - 0.5x$, where $p$ is the price per unit, and the cost function is $C(x) = 10000 + 50x + 0.1x^2$. Which production level $x$ maximizes profit?
    Answer: $375$ The revenue function is $R(x) = x \cdot p(x) = x(500 - 0.5x) = 500x - 0.5x^2$. The profit function is $P(x) = R(x) - C(x) = (500x - 0.5x^2) - (10000 + 50x + 0.1x^2) = -0.6x^2 + 450x - 10000$. To find the maximum profit, we differentiate $P(x)$ and set it to zero: $P'(x) = -1.2x + 450 = 0$. Solving for $x$ gives $1.2x = 450$, so $x = 450/1.2 = 4500/12 = 375$. The second derivative is $P''(x) = -1.2$, which is negative, confirming that $x=375$ yields a maximum profit.
18. A rectangle is constructed with its base on the x-axis and its two upper vertices on the parabola $y = 12 - x^2$. What is the maximum possible perimeter of such a rectangle?
    Answer: 26 Let the upper right vertex of the rectangle be at $(x, y)$, where $y = 12 - x^2$. The width of the rectangle is $2x$ and the height is $y = 12 - x^2$. The perimeter $P$ is given by $P(x) = 2(2x) + 2(12 - x^2) = 4x + 24 - 2x^2$, for $x$ in $[0, \sqrt{12}]$. To find the maximum perimeter, we take the derivative: $P'(x) = 4 - 4x$. Setting $P'(x) = 0$ gives $4 - 4x = 0$, so $x = 1$. To find the maximum value, we check the critical point and the endpoints: $P(1) = 4(1) + 24 - 2(1)^2 = 26$. $P(0) = 24$. $P(\sqrt{12}) = 4\sqrt{12} = 8\sqrt{3} \approx 13.86$. The maximum value is 26. A: This is the maximum area of the rectangle, found by maximizing $A(x) = 2x(12-x^2)$, which occurs at $x=2$, giving $A(2)=32$. This is a common error of optimizing the wrong quantity. C: This is the perimeter of the rectangle when its area is maximized (at $x=2$), $P(2) = 24$. It is also the perimeter at the endpoint $x=0$. D: This is the height (11) plus the width (2) plus a small value, or possibly an arithmetic error. A rectangle of perimeter 16 would have dimensions like $x=2, y=6$, which would be $P=2(4)+2(6)=20$, so this is likely just a calculation error.
19. What is the x-coordinate of the point on the parabola $y=x^2$ that is closest to the point $(18, 0)$?
    Answer: $2$ Let a point on the parabola be $(x, x^2)$. The distance $D$ between this point and $(18, 0)$ is given by $D = \sqrt{(x-18)^2 + (x^2-0)^2}$. Minimizing $D$ is equivalent to minimizing its square, $S = D^2 = (x-18)^2 + (x^2)^2 = x^2 - 36x + 324 + x^4$. To find the minimum, we differentiate $S$ with respect to $x$: $S'(x) = 2x - 36 + 4x^3$. Setting $S'(x) = 0$ gives $4x^3 + 2x - 36 = 0$, or $2x^3 + x - 18 = 0$. By the Rational Root Theorem, we can test integer factors of 18. Testing $x=2$, we get $2(2^3) + 2 - 18 = 2(8) + 2 - 18 = 16 + 2 - 18 = 0$. So, $x=2$ is the critical point. The second derivative is $S''(x) = 12x^2 + 2$, which is positive for all $x$, confirming that $x=2$ corresponds to a minimum distance.
20. The function $f(x) = x^4 - 4x^3 + 10$ has local extrema. What is the absolute minimum value of the function on the interval $[-1, 4]$?
    Answer: -17 To find the absolute minimum on a closed interval, we use the Closed Interval Method. First, find the critical points by taking the derivative: $f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$. Setting $f'(x) = 0$ gives critical points at $x=0$ and $x=3$. Both are in the interval $[-1, 4]$. Now, we evaluate the function at the critical points and the endpoints: $f(-1) = (-1)^4 - 4(-1)^3 + 10 = 1 - 4(-1) + 10 = 1 + 4 + 10 = 15$. $f(0) = (0)^4 - 4(0)^3 + 10 = 10$. $f(3) = (3)^4 - 4(3)^3 + 10 = 81 - 4(27) + 10 = 81 - 108 + 10 = -17$. $f(4) = (4)^4 - 4(4)^3 + 10 = 256 - 4(64) + 10 = 256 - 256 + 10 = 10$. Comparing the values ${15, 10, -17, 10}$, the absolute minimum is -17. A: Correct. B: This is the value at the critical point $x=0$ and the endpoint $x=4$. It is a local maximum or inflection point value, but not the absolute minimum. C: This is the value at the endpoint $x=-1$, which is the absolute maximum on the interval. D: This is the value of $f(3)$ with a sign error in calculation: $81-108+10 = -17$. Maybe $81-108 = -27$ and the student forgot to add 10.

Start this quiz