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.

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.

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.

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.

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.

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.

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.

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.

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.

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.