Solving Optimization Problems - AP Calculus BC

A farmer has $100$ m of fencing for three sides of a rectangle against a wall. With width $x$ , area is $A(x)=x(100-2x)$ . Which $x$ maximizes $A$ ?
Answer: $25$ This optimization problem involves maximizing the area of a rectangle with one side against a wall. Given A(x) = x(100-2x), we find A'(x) = 100 - 4x. Setting A'(x) = 0 gives 100 - 4x = 0, so x = 25. Since A''(x) = -4 < 0, this critical point is a maximum. The constraint 0 < x < 50 (since 100-2x > 0) is satisfied. A tempting wrong answer is x = 50 (option D), which would use all the fencing for the width but leave no fencing for the length parallel to the wall. The key optimization principle is to express the objective function in terms of one variable, find critical points, and verify they satisfy all constraints.
A rectangle has base on the $x$ -axis and upper corners on $y=8-2x^2$ . If right upper corner is at $x$ , what $x$ maximizes area?
Answer: $\sqrt{\frac{4}{3}}$ 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.
A farmer has $100$ meters of fencing for three sides of a rectangle against a wall; what width $x$ maximizes area?
Answer: $25$ 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 rectangular enclosure with 100 meters of fencing for three sides against a wall, with width x, the length is 100 - 2x, so area A(x) = x(100 - 2x). To maximize A(x), compute A'(x) = 100 - 4x = 0, giving critical point x = 25. The second derivative A''(x) = -4 < 0 confirms a maximum; endpoints x = 0 and x = 50 give A = 0. A tempting distractor is x = 50, 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.
For $x>0$ , the cost is $C(x)=x+\frac{36}{x}$ ; which $x$ minimizes the cost?
Answer: $6$ This problem requires solving an optimization problem to minimize a cost function. Given C(x) = x + 36/x for x > 0, we find critical points by taking the derivative: C'(x) = 1 - 36/x². Setting C'(x) = 0: 1 = 36/x², so x² = 36, giving x = 6 (since x > 0). To verify this is a minimum, C''(x) = 72/x³ > 0 for all x > 0, confirming x = 6 minimizes cost. At x = 6, C(6) = 6 + 6 = 12. Students might choose x = 4, but C(4) = 4 + 9 = 13 > 12. For functions of the form f(x) = x + k/x, the minimum occurs at x = √k—a useful pattern to remember.
A farmer has $100$ meters of fencing for three sides of a rectangle against a river; which width $x$ maximizes area?
Answer: $25$ 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 the area with 100 meters of fencing for three sides against a river and width x, express area as A(x)=x(100-2x) and find the critical point by setting A'(x)=100-4x=0, yielding x=25. This critical point is a maximum because the second derivative A''(x)=-4 is negative. Evaluating at endpoints x=0 and x=50 gives A=0, confirming the maximum at x=25. A tempting distractor is x=50, but that gives zero area as the length becomes zero. 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.
A wire of length $24$ is cut into two pieces forming squares; if first piece length is $x$ , which $x$ minimizes total area?
Answer: $12$ 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.
A rectangle has sides $x$ and $12-2x$ for $0 \leq x \leq 6$ . What $x$ maximizes area?
Answer: $3$ This problem requires solving an optimization problem using calculus to maximize the area of a rectangle with sides $x$ and $12-2x$ for $0 \leq x \leq 6$ . The area function is $A(x) = x(12-2x) = 12x - 2x^2$ , and its derivative $A'(x) = 12 - 4x$ sets to zero at $x = 3$ . Evaluating at the critical point $x = 3$ and endpoints $x = 0$ and $x = 6$ shows $A(3) = 18$ , while $A(0) = 0$ and $A(6) = 0$ , so the maximum is at $x = 3$ . This critical point is confirmed as a maximum since $A''(x) = -4 < 0$ . A tempting distractor is $x = 4$ , but it fails because $A(4) = 4 \times 4 = 16 < 18$ , not the maximum. 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.
A company's profit is $P(x)=-x^2+10x-9$ for $0 \le x \le 10$ . What $x$ maximizes profit?
Answer: $5$ 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.
A cylindrical can has fixed volume $V=16\pi$ . With radius $r>0$ , surface area is $S(r)=2\pi r^2+\frac{32\pi}{r}$ . Which $r$ minimizes $S$ ?
Answer: $2$ This problem involves minimizing the surface area of a cylinder with fixed volume. Given S(r) = 2πr² + 32π/r, we find S'(r) = 4πr - 32π/r². Setting S'(r) = 0 gives 4πr - 32π/r² = 0, so 4r³ = 32, yielding r³ = 8 and r = 2. To confirm this is a minimum, S''(r) = 4π + 64π/r³, and S''(2) = 4π + 64π/8 = 12π > 0. A common error is choosing r = ∛4 (option C) by mishandling the algebra when solving r³ = 8. The optimization principle for constrained problems is to express the objective function in terms of one variable using the constraint, then find critical points through standard calculus techniques.
A rectangle has fixed area $36$ ; if one side is $x$ , what $x$ minimizes the sum of side lengths $x+\frac{36}{x}$ ?
Answer: $6$ 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.
For $x>0$ , minimize $f(x)=4x+\frac{9}{x}$ . What value of $x$ minimizes $f$ ?
Answer: $\frac{3}{2}$ This problem requires solving an optimization problem using calculus to minimize $f(x) = 4x + \frac{9}{x}$ for $x > 0$ . The derivative $f'(x) = 4 - \frac{9}{x^2}$ sets to zero at $x = \frac{3}{2}$ . With no endpoints, this is the minimum as $f''(x) > 0$ there. Evaluation gives the minimum value of $12$ . A tempting distractor is $\frac{9}{4}$ , but it fails because it does not solve the equation and gives higher value. 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.
For $0\le x\le 6$ , a revenue function is $R(x)=x(6-x)$ . What $x$ maximizes revenue?
Answer: $3$ 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.
Two sides of a rectangle lie along axes; the corner $(x,y)$ is on $x+2y=12$ ; which $x$ maximizes area?
Answer: $6$ 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 the area of the rectangle with corner (x,y) on $x+2y=12$ , express area as $A(x)=x(6 - x/2)$ and find the critical point by setting $A'(x)=6 - x=0$ , yielding $x=6$ . This critical point is a maximum because the second derivative $A''(x)=-1$ is negative. Evaluating at endpoints $x=0$ and $x=12$ gives $A=0$ , confirming the maximum at $x=6$ . A tempting distractor is $x=12$ , but that gives zero area, which is the minimum. 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.
For $x>0$ , the cost per unit is $C(x)=x+\frac{36}{x}$ . What $x$ minimizes $C(x)$ ?
Answer: $6$ 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.
A ball is thrown upward; its height is $h(t)=-16t^2+64t$ . For $t\ge0$ , at what $t$ is height maximized?
Answer: $2$ 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 height h(t) = -16t² + 64t for t ≥ 0, it is a quadratic opening downward. To maximize h(t), compute h'(t) = -32t + 64 = 0, giving critical point t = 2. At t = 0, h = 0, and as t increases beyond, h decreases after t = 2. A tempting distractor is t = 4, but it gives lower height than at t = 2. 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.
A right circular cylinder has fixed surface area $S=2\pi r^2+2\pi rh=54\pi$ . What radius $r$ maximizes volume?
Answer: $3$ 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.
A projectile's height is $h(t)=-t^2+6t$ for $0 \le t \le 6$ ; at what time $t$ is height maximized?
Answer: $3$ 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.
For $0<x<4$ , the cost is $C(x)=x^3-6x^2+12x+5$ . Which $x$ minimizes $C$ ?
Answer: $2$ This problem requires finding the minimum cost by solving an optimization problem. For C(x) = x³ - 6x² + 12x + 5, we compute C'(x) = 3x² - 12x + 12 = 3(x² - 4x + 4) = 3(x-2)². Setting C'(x) = 0 gives x = 2. To verify this is a minimum, we check C''(x) = 6x - 12, so C''(2) = 0, requiring the second derivative test to be inconclusive. However, since C'(x) = 3(x-2)² ≥ 0 with equality only at x = 2, the function is increasing on both sides of x = 2, confirming it's a minimum. Students might choose x = 0 (option A) thinking the minimum cost occurs with no production. The strategy for optimization is to find where the derivative equals zero and verify the nature of critical points using either the second derivative test or by analyzing the sign of the first derivative.
For $0<x<10$ , which $x$ maximizes $P(x)=x\sqrt{10-x}$ , the profit from producing $x$ units?
Answer: $20/3$ Solving optimization problems is a key skill in AP Calculus BC, involving finding maximum or minimum values of a function subject to constraints. For profit P(x) = x√(10 - x) on (0, 10), set P'(x) = (20 - 3x)/(2√(10 - x)) = 0, yielding x = 20/3. Evaluate at the critical point and note values approach 0 at endpoints. The first derivative test confirms a maximum at x = 20/3. A tempting distractor is x = 5, but P(5) < P(20/3). Always remember to check both critical points and endpoints in closed intervals for optimization problems.
Find the value of $x$ on $[0,6]$ that maximizes $f(x)=x(6-x)^2$ .
Answer: $2$ 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.

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