This problem requires solving optimization problems using calculus, a key skill in AP Calculus AB. To maximize the area A = $x(20 - x)$, compute the derivative A' = $20 - 2x$ and set it to zero, yielding a critical point at $x = 10$. Evaluating at $x = 10$ gives the maximum area, while at the endpoints $x = 0$ and $x = 20$, A = $0$. The second derivative A'' = $-2$ is negative, confirming a maximum. A tempting distractor is $x = 0$, but it yields zero area, which is the minimum, not the maximum. To solve optimization problems, express the quantity as a function of one variable, find critical points by setting the derivative to zero, and evaluate at critical points and endpoints to find the maximum or minimum.

This problem requires solving optimization problems using calculus, a key skill in AP Calculus AB. To maximize the area A = x(18 - x), compute the derivative A' = 18 - 2x and set it to zero, yielding a critical point at x = 9. Evaluating at x = 9 gives the maximum, while endpoints yield zero. The second derivative A'' = -2 is negative, confirming a maximum. A tempting distractor is x = 6, but it yields a smaller area as it is not the critical point. To solve optimization problems, express the quantity as a function of one variable, find critical points by setting the derivative to zero, and evaluate at critical points and endpoints to find the maximum or minimum.

This problem requires solving optimization problems using calculus, a key skill in AP Calculus AB. To maximize the area A = x(200 - 2x), compute the derivative A' = 200 - 4x and set it to zero, yielding a critical point at x = 50. Evaluating at x = 50 gives the maximum area, while at the endpoints x = 0 and x = 100, A = 0. The second derivative A'' = -4 is negative, confirming a maximum. A tempting distractor is x = 100, but it yields zero area as the width becomes zero. To solve optimization problems, express the quantity as a function of one variable, find critical points by setting the derivative to zero, and evaluate at critical points and endpoints to find the maximum or minimum.

This problem requires solving optimization problems using calculus, a key skill in AP Calculus AB. To minimize the perimeter $P = 2\left(x + \frac{36}{x}\right)$, compute the derivative $P' = 2\left(1 - \frac{36}{x^2}\right)$ and set it to zero, yielding $x = 6$. Evaluating confirms this as the minimum, with second derivative positive. Endpoints are not applicable as $x > 0$. A tempting distractor is $x = 12$, but it yields a higher perimeter. To solve optimization problems, express the quantity as a function of one variable, find critical points by setting the derivative to zero, and evaluate at critical points and endpoints to find the maximum or minimum.

This problem involves solving optimization problems by maximizing f(x) = x √(9 - x) for 0 < x < 9. To find the maximum, compute the derivative f'(x) = √(9 - x) - x/(2√(9 - x)) and set it to zero, yielding x = 6 as the critical point. Evaluating f at x = 6 gives the maximum of 6√3 ≈ 10.39. Since the interval is open, endpoints are not evaluated, but give limits of 0. A tempting distractor is x = 3, where f=3√6 ≈ 7.35, less than the maximum. A transferable optimization-solving strategy is to identify critical points by setting the derivative to zero, evaluate the function at those points and any endpoints if applicable, and compare values to determine the extreme.

This problem requires solving optimization problems using calculus, a key skill in AP Calculus AB. To maximize the volume V = 4x(10 - x)², compute the derivative V' = 4(10 - x)(10 - 3x) and set it to zero, yielding x = 10/3. Evaluating at x = 10/3 gives the maximum, while endpoints yield zero. The second derivative confirms a maximum. A tempting distractor is x = 5, but it yields a lower volume. To solve optimization problems, express the quantity as a function of one variable, find critical points by setting the derivative to zero, and evaluate at critical points and endpoints to find the maximum or minimum.

This problem involves solving optimization problems by maximizing the area A = 2x(9 - x²) of a rectangle under y=9-x². To find the maximum, compute the derivative A'(x) = 2(9 - x²) - 4x² = 18 - 6x² and set it to zero, yielding x = √3 as the critical point. Evaluating A at x = √3 gives the maximum. The interval is implicit for x > 0, and no endpoints are needed. A tempting distractor is x = 3, where area is zero, but this is outside the domain or a minimum. A transferable optimization-solving strategy is to identify critical points by setting the derivative to zero, evaluate the function at those points and any endpoints if applicable, and compare values to determine the extreme.

This problem involves solving an optimization problem to maximize the area of a rectangular pen. Given A(x) = x(200-2x), we find the maximum by taking the derivative: A'(x) = 200 - 4x. Setting A'(x) = 0 gives 200 - 4x = 0, which solves to x = 50. To verify this is a maximum, we check the second derivative: A''(x) = -4 < 0, confirming a maximum at x = 50. The domain constraint requires x > 0 and 200 - 2x > 0, giving 0 < x < 100, so x = 50 is valid. Students might choose x = 100 thinking it maximizes one dimension, but this would make the other dimension zero. The optimization strategy is to express the objective function in one variable, find critical points using calculus, and verify the solution satisfies all constraints.

This fence optimization problem requires maximizing area with a perimeter constraint. The area function A(x) = x(200-2x) = 200x - 2x² has derivative A'(x) = 200 - 4x. Setting A'(x) = 0 gives 200 - 4x = 0, so x = 50. Since A''(x) = -4 < 0, this critical point maximizes the area. The choice x = 100 would use all fencing for width, leaving no fencing for the length, resulting in zero area. For rectangular optimization problems with one side against a wall, the optimal width is typically one-fourth of the total fencing available.

This problem seeks a local minimum of a cubic function. For f(x) = x³ - 12x + 5, we find f'(x) = 3x² - 12. Setting f'(x) = 0 gives 3x² - 12 = 0, so x² = 4, which means x = ±2. Since we need x > 0, we consider x = 2. To verify it's a minimum, we check f''(x) = 6x, so f''(2) = 12 > 0, confirming x = 2 gives a local minimum. The choice x = -2 would give a local maximum, not minimum. When finding extrema of cubic functions, use the second derivative test to distinguish between maxima and minima.