AP Calculus BC Quiz: Solving Optimization Problems

What this quiz covers

This quiz focuses on Solving Optimization Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $0$
2. $3$ (correct answer)
3. $6$
4. $2$
5. $4$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. For height $h(t) = -t^2 + 6t$ on $[0, 6]$, set $h'(t) = -2t + 6 = 0$, giving $t = 3$. Evaluate $h(3) = 9$ compared to $h(0) = 0$ and $h(6) = 0$. The negative second derivative confirms a maximum at $t = 3$. A tempting distractor is $t = 6$, but $h(6) = 0$ is the minimum. Always remember to check both critical points and endpoints in closed intervals for optimization problems.

Question 2

1. $0$
2. $6$
3. $2$
4. $3$ (correct answer)
5. $4$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. For the area A(x) = x(6 - x) on [0, 6], find critical points by setting A'(x) = 6 - 2x = 0, yielding x = 3. Evaluate at the critical point where A(3) = 9 and at endpoints A(0) = 0 and A(6) = 0. This confirms the maximum at x = 3 since the function is a downward parabola. A tempting distractor is x = 4, but it gives A = 8, which is less than 9. Always remember to check both critical points and endpoints in closed intervals for optimization problems.

Question 3

1. $25$
2. $5$ (correct answer)
3. $1$
4. rac{1}{5}
5. $0$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. For $f(x) = \frac{25}{x} + x$ for $x > 0$, compute $f'(x) = -\frac{25}{x^2} + 1 = 0$, giving $x = 5$. The second derivative $f''(x) = \frac{50}{x^3} > 0$ for $x > 0$ confirms a minimum. As $x$ approaches 0 or infinity, $f$ increases. A tempting distractor is $x = 25$, but it gives higher value than at $x = 5$. A transferable optimization-solving strategy is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, evaluate at critical points and endpoints, and use tests to confirm the extremum.

Question 4

1. $3$
2. $6$ (correct answer)
3. $9$
4. $12$
5. $18$

Explanation: This problem requires solving an optimization problem using calculus to minimize the sum of side lengths x + 36/x for a rectangle with fixed area 36. The sum function is S(x) = x + 36/x, and its derivative S'(x) = 1 - 36/x² sets to zero at x = 6. Assuming x > 0 with no endpoints, this critical point is the minimum as S''(x) > 0 there. Evaluating shows the minimum sum is 12 at x = 6. A tempting distractor is x = 9, but it fails because S(9) = 9 + 4 = 13 > 12. A transferable strategy for optimization is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, and check endpoints if the domain is closed.

Question 5

1. $3$ (correct answer)
2. $6$
3. $root{18}$
4. $root{9}$
5. $2$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. For the cylinder with fixed surface area 54π, express h = (27 - r²)/r, volume V(r) = π(27r - r³). To maximize V(r), set V'(r) = π(27 - 3r²) = 0, giving r = 3. The second derivative test confirms a maximum. A tempting distractor is r = 6, but it gives lower volume. A transferable optimization-solving strategy is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, evaluate at critical points and endpoints, and use tests to confirm the extremum.

Question 6

1. $1$
2. $\sqrt{\frac{4}{3}}$ (correct answer)
3. $\sqrt{2}$
4. $2$
5. $0$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. For the rectangle with base on x-axis and upper corners on $y = 8 - 2x^2$, with right upper at $x$, assuming symmetry, $A(x) = 2x(8 - 2x^2)$. To maximize $A(x)$, set $A'(x) = 16 - 12x^2 = 0$, giving $x = \sqrt{\frac{4}{3}}$. Endpoints give zero area. A tempting distractor is $x = \sqrt{2}$, but it gives lower area. A transferable optimization-solving strategy is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, evaluate at critical points and endpoints, and use tests to confirm the extremum.

Question 7

1. $12$
2. $0$
3. $6$ (correct answer)
4. $4$
5. $8$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. For the triangle with base x and height 12 - x on [0, 12], area A(x) = (1/2)x(12 - x). To maximize A(x), set A'(x) = (1/2)(12 - 2x) = 0, giving x = 6. Endpoints A(0) = 0 and A(12) = 0 confirm the maximum. A tempting distractor is x = 12, but it gives zero area. A transferable optimization-solving strategy is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, evaluate at critical points and endpoints, and use tests to confirm the extremum.

Question 8

1. $0$
2. $8$
3. $12$ (correct answer)
4. $16$
5. $24$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. To minimize the total area of two squares from a 24-unit wire cut at x, express A(x)=[x² + (24-x)²]/16 and find the critical point by setting A'(x)=(1/4)x - 3=0, yielding x=12. This critical point is a minimum because the second derivative A''(x)=1/4 is positive. Evaluating at endpoints x=0 and x=24 gives A=36, while A(12)=18 is smaller. A tempting distractor is x=0, but that gives a larger area of 36 compared to 18 at x=12. A transferable strategy for optimization is to express the quantity to optimize as a function of one variable, find critical points by setting the derivative to zero, and use the second derivative test or endpoint evaluation to identify maxima or minima.

Question 9

1. $0$
2. $2$ (correct answer)
3. $3$
4. $4$
5. $6$

Explanation: This problem requires solving an optimization problem to find the production level that maximizes profit. Expanding P(x) = x(6-x)² = x(36 - 12x + x²) = 36x - 12x² + x³. Taking the derivative: P'(x) = 36 - 24x + 3x² = 3(x² - 8x + 12) = 3(x-2)(x-6). Critical points occur at x = 2 and x = 6. Evaluating: P(0) = 0, P(2) = 2(4)² = 32, P(6) = 6(0)² = 0. Therefore, x = 2 gives the maximum profit of 32. Students might choose x = 3 (the midpoint), but this gives P(3) = 3(3)² = 27 < 32. The strategy is to find all critical points and evaluate at both critical points and endpoints to identify the global maximum.

Question 10

1. $0$
2. $3$
3. $6$ (correct answer)
4. $9$
5. $12$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. To maximize profit $P(x)=-x^2+12x-20$, find the critical point by setting $P'(x)=-2x+12=0$, yielding $x=6$. This critical point is a maximum because the second derivative $P''(x)=-2$ is negative. No specific endpoints are given, but the parabola opens downward, confirming the vertex is the maximum. A tempting distractor is $x=12$, but that gives $P=-20$, which is less than 16 at $x=6$. A transferable strategy for optimization is to express the quantity to optimize as a function of one variable, find critical points by setting the derivative to zero, and use the second derivative test or endpoint evaluation to identify maxima or minima.

Question 11

1. $1$
2. $2$
3. $3$ (correct answer)
4. $6$
5. $0$

Explanation: This problem requires solving an optimization problem using calculus to maximize the area of a rectangle under $y=6-x$ in the first quadrant with one corner at the origin. The area $A(x) = x(6 - x)$, and $A'(x) = 6 - 2x$ sets to zero at $x = 3$. Evaluating at $x = 3$ and endpoint $x = 0$ confirms the maximum. The second derivative is negative at $x = 3$. A tempting distractor is $x = 2$, but it fails because $A(2) = 8 < 9$ at $x = 3$. A transferable strategy for optimization is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, and check endpoints if the domain is closed.

Question 12

1. $3$
2. $6$ (correct answer)
3. $12$
4. $18$
5. $0$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. For the cost $C(x) = x + \frac{36}{x}$ for $x > 0$, compute the derivative $C'(x) = 1 - \frac{36}{x^2} = 0$, giving $x = 6$. The second derivative $C''(x) = \frac{72}{x^3} > 0$ for $x > 0$ confirms a minimum. As $x$ approaches 0 or infinity, $C$ increases, supporting the minimum at $x = 6$. A tempting distractor is $x = 12$, but it gives higher cost than at $x = 6$. A transferable optimization-solving strategy is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, evaluate at critical points and endpoints, and use tests to confirm the extremum.

Question 13

1. $10$
2. $5$
3. $\frac{10}{\sqrt{2}}$ (correct answer)
4. $\sqrt{10}$
5. $0$

Explanation: This problem requires solving an optimization problem using calculus to maximize the area of a rectangle with fixed diagonal 10 and one side x. The area is A(x) = x $\sqrt{100 - x^2}$, and the derivative A'(x) = $\frac{100 - 2x^2}{\sqrt{100 - x^2}}$ sets to zero when $100 - 2x^2 = 0$, so x = $\frac{10}{\sqrt{2}}$. Assuming domain 0 < x < 10, there are no endpoints, but checking confirms this is the maximum as it forms a square. The second derivative or sign change verifies the maximum. A tempting distractor is $\sqrt{10}$, but it fails because it does not solve the derivative equation and gives a lower area. A transferable strategy for optimization is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, and check endpoints if the domain is closed.

Question 14

1. $5$
2. $10$ (correct answer)
3. $20$
4. $15$
5. $0$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. For the rectangle with fixed perimeter 40 and width x, the length is 20 - x, so the area A(x) = x(20 - x) = 20x - x². To maximize A(x), compute the derivative A'(x) = 20 - 2x and set it to zero, giving the critical point x = 10. The second derivative A''(x) = -2 < 0 confirms a maximum at x = 10; checking endpoints x = 0 and x = 20 yields A = 0, the minimum. A tempting distractor is x = 20, but it results in zero area as the length becomes zero. A transferable optimization-solving strategy is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, evaluate at critical points and endpoints, and use tests to confirm the extremum.

Question 15

1. $12$
2. $0$
3. $4$
4. $6$ (correct answer)
5. $8$

Explanation: This problem requires solving an optimization problem using calculus to maximize the triangle area A = (1/2) x (12 - x) for 0 ≤ x ≤ 12. The derivative A'(x) = (1/2)(12 - 2x) sets to zero at x = 6. Evaluating at x = 6 and endpoints x = 0 and x = 12 shows A(6) = 18 is the maximum. The second derivative is negative, confirming the maximum. A tempting distractor is x = 8, but it fails because A(8) = (1/2)84 = 16 < 18. A transferable strategy for optimization is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, and check endpoints if the domain is closed.

Question 16

1. $10$
2. $0$
3. $4$
4. $5$ (correct answer)
5. $9$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. For the profit $P(x) = -x^2 + 10x - 9$ on $[0, 10]$, it is a quadratic opening downward. To maximize P(x), compute $P'(x) = -2x + 10 = 0$, giving critical point $x = 5$. Evaluating at endpoints $P(0) = -9$, $P(10) = -9$, and $P(5) = 16$ confirms the maximum. A tempting distractor is $x = 10$, but it gives the same low profit as $x = 0$. A transferable optimization-solving strategy is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, evaluate at critical points and endpoints, and use tests to confirm the extremum.

Question 17

1. $0$
2. $2$ (correct answer)
3. $3$
4. $4$
5. $6$

Explanation: This problem asks us to solve an optimization problem by finding the maximum of f(x) = x(6-x)² on the interval [0,6]. To find critical points, we compute f'(x) = (6-x)² + x·2(6-x)(-1) = (6-x)² - 2x(6-x) = (6-x)[(6-x) - 2x] = (6-x)(6-3x). Setting f'(x) = 0 gives x = 6 or x = 2. We must check the critical points x = 2, x = 6 and the endpoint x = 0: f(0) = 0, f(2) = 2(4)² = 32, and f(6) = 6(0)² = 0. The tempting answer x = 3 might come from incorrectly setting only one factor to zero or confusing the arithmetic. When optimizing on a closed interval, always evaluate the function at critical points AND endpoints to find the global maximum.

Question 18

1. $10$
2. $20$
3. $25$ (correct answer)
4. $50$
5. $100$

Explanation: This problem involves solving an optimization problem where a farmer uses 100 meters of fencing for three sides of a rectangle against a wall. If the width perpendicular to the wall is x, then the two sides perpendicular to the wall use 2x meters, leaving 100-2x meters for the side parallel to the wall. The area is A(x) = x(100-2x) = 100x - 2x². Taking the derivative: A'(x) = 100 - 4x, and setting it to zero gives x = 25. Since A''(x) = -4 < 0, this is indeed a maximum. The answer x = 50 might tempt students who forget to account for two sides of width x, thinking only 50 meters remains for the parallel side. In constrained optimization problems, carefully translate the physical constraints into mathematical relationships before differentiating.

Question 19

1. $0$
2. $6$
3. $3$ (correct answer)
4. $2$
5. $4$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. For the revenue function R(x) = x(6 - x) on [0, 6], it is a quadratic opening downward. To maximize R(x), compute R'(x) = 6 - 2x = 0, giving critical point x = 3. Evaluating at endpoints R(0) = 0 and R(6) = 0, and R(3) = 9 confirms the maximum. A tempting distractor is x = 6, but it gives zero revenue, not the maximum. A transferable optimization-solving strategy is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, evaluate at critical points and endpoints, and use tests to confirm the extremum.

Question 20

1. $20$
2. $0$
3. $5$
4. $10$ (correct answer)
5. $15$

Explanation: Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. For the revenue R(x) = 20x - x² on [0, 20], it is a quadratic opening downward. To maximize R(x), set R'(x) = 20 - 2x = 0, giving x = 10. Endpoints R(0) = 0 and R(20) = 0 confirm the maximum. A tempting distractor is x = 20, but it gives zero revenue. A transferable optimization-solving strategy is to express the quantity as a function of one variable, find critical points by setting the derivative to zero, evaluate at critical points and endpoints, and use tests to confirm the extremum.