This problem requires solving an optimization problem to find the production level that maximizes profit. For $P(x) = -2x^2 + 40x - 90$, we find the maximum by taking the derivative: $P'(x) = -4x + 40$. Setting $P'(x) = 0$ gives $-4x + 40 = 0$, which solves to $x = 10$. Since $P''(x) = -4 < 0$, this confirms $x = 10$ gives a maximum. The parabola opens downward (negative leading coefficient), so the vertex represents the maximum profit point. Students might incorrectly choose $x = 20$ by misreading the coefficient or $x = 40$ by solving $-x + 40 = 0$ instead. The key optimization principle is that for quadratic functions, the maximum or minimum occurs at the vertex, found where the derivative equals zero.

This problem requires solving an optimization problem to maximize revenue. Given R(x) = 120x - 3x², we find the maximum by taking the derivative: R'(x) = 120 - 6x. Setting R'(x) = 0 gives 120 - 6x = 0, which solves to x = 20. Since R''(x) = -6 < 0, this confirms x = 20 gives a maximum. We verify this satisfies the constraint 0 ≤ x ≤ 40. Students might incorrectly choose x = 40 thinking the endpoint maximizes revenue, but R(40) = 3600 < R(20) = 1200. The key optimization strategy is to check both critical points and endpoints when dealing with closed intervals, selecting the point that gives the largest function value.

This triangle area optimization is similar to the first problem. To maximize A(x) = ½x(12-x) = 6x - ½x², we find A'(x) = 6 - x. Setting A'(x) = 0 gives 6 - x = 0, so x = 6. Since A''(x) = -1 < 0, this confirms x = 6 maximizes the area. The choice x = 12 would make one leg zero, resulting in zero area. For right triangle optimization problems where the sum of legs is fixed, the maximum area occurs when the legs are equal, giving an isosceles right triangle.

This problem requires solving an optimization problem to maximize revenue. Given R(x) = 40x - x², we find R'(x) = 40 - 2x = 0, which gives x = 20. Since R''(x) = -2 < 0, this is a maximum. We should also check endpoints: R(0) = 0 and R(40) = 1600 - 1600 = 0. Comparing R(20) = 800 - 400 = 400 with the endpoint values confirms x = 20 maximizes revenue. The answer x = 40 might seem appealing as the upper bound, but it gives zero revenue. For quadratic optimization problems, the vertex formula or calculus both lead to the same optimal solution.

This problem involves solving optimization problems by maximizing A(x) = √x (16 - x) for 0 < x < 16. To find the maximum, compute the derivative A'(x) = 8/√x - (3/2)√x and set it to zero, yielding x = 16/3 as the critical point. Evaluating A at x = 16/3 gives the maximum. Since the interval is open, endpoints are not evaluated, but the critical point provides the solution. A tempting distractor is x = 8, where A is less than at x=16/3 due to not balancing the terms optimally. 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 revenue R(x) = x(30 - x), compute the derivative R' = 30 - 2x and set it to zero, yielding a critical point at x = 15. Evaluating at x = 15 gives the maximum, while endpoints yield zero. The second derivative R'' = -2 is negative, confirming a maximum. A tempting distractor is x = 0, but it yields zero revenue, which is a minimum. 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 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 requires solving optimization problems using calculus, a key skill in AP Calculus AB. To maximize $f(x) = x(5 - x)^2$, compute the derivative $f' = (5 - x)(5 - 3x)$ and set it to zero, yielding $x = \frac{5}{3}$. Evaluating at $x = \frac{5}{3}$ gives the maximum, while endpoints yield zero. The second derivative confirms a maximum. A tempting distractor is $x = \frac{5}{2}$, but it yields a lower value 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 involves solving optimization problems by maximizing f(x) = sin x - x/10 on 0 < x < π. To find the maximum, compute the derivative f'(x) = cos x - 1/10 and set it to zero, yielding x = arccos(1/10) as the critical point. Evaluating f at this point confirms the maximum by the second derivative test. Since the interval is open, endpoints are not evaluated, but the critical point provides the solution. A tempting distractor is x = π/2, where f is less due to the linear term dominating. 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 optimization problems by finding the minimum value of f(x) = x + 9/x on 0 < x < 6. To find the minimum, compute the derivative f'(x) = 1 - 9/x² and set it to zero, yielding x = 3 as the critical point. Evaluating f at x = 3 gives the minimum, confirmed by the second derivative test showing f''(3) > 0. Since the interval is open, endpoints are not evaluated, but the critical point provides the minimum. A tempting distractor is x = 6, where f(x) = 7.5, but this is higher than f(3) = 6 and near the boundary. 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.