Solving Optimization Problems - AP Calculus AB

A rectangle's length is $x$ and width is $\frac{36}{x}$ ; which $x$ minimizes perimeter $P=2\left(x+\frac{36}{x}\right)$ ?
Answer: $x=6$ 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.
A rectangle has one vertex at the origin and opposite vertex on $y=16-2x^2$ in the first quadrant; if $x$ is the $x$ -coordinate, which $x$ maximizes area $A=x(16-2x^2)$ ?
Answer: $x=\sqrt{\frac{8}{3}}$ This problem requires solving optimization problems using calculus, a key skill in AP Calculus AB. To maximize A = x(16 - 2x²), compute the derivative A' = 16 - 6x² and set it to zero, yielding x = √(8/3). Evaluating at x = √(8/3) gives the maximum, while endpoints yield zero. The second derivative A'' = -12x is negative for x > 0, confirming a maximum. A tempting distractor is x = √8, but it yields a smaller area as it ignores the coefficient. 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.
A rectangle with perimeter $60$ has length $x$ and width $30-x$ ; which $x$ maximizes area $A=x(30-x)$ ?
Answer: $x=15$ This problem involves solving optimization problems by maximizing the area $A(x) = x(30 - x)$ for a rectangle with perimeter 60. To find the maximum, compute the derivative $A'(x) = 30 - 2x$ and set it to zero, yielding $x = 15$ as the critical point. Evaluating $A$ at $x = 15$ gives the maximum area of 225. The problem implies a closed interval, but the critical point determines the solution without needing endpoints. A tempting distractor is $x = 30$ , where area is zero, but this is a minimum at 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.
A rancher has 200 ft of fencing for three sides of a rectangle against a wall. Which width $x$ maximizes area $A(x)=x(200-2x)$ ?
Answer: $x=50$ 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.
A projectile's height is $h(t)=-16t^2+64t$ for $0\le t\le 4$ . At what time $t$ is $h(t)$ maximized?
Answer: $t=2$ This projectile motion problem requires finding when height is maximized. For h(t) = -16t² + 64t, we find h'(t) = -32t + 64. Setting h'(t) = 0 gives -32t + 64 = 0, so t = 2. Since h''(t) = -32 < 0, this confirms t = 2 gives the maximum height. The endpoint t = 4 is tempting since it's when the projectile returns to ground level, but the maximum occurs at the vertex of the parabola. For quadratic motion problems, the maximum always occurs at the critical point where the velocity (first derivative) equals zero.
A farmer fences a rectangular pen against a wall using $200$ m of fencing; which width $x$ maximizes area $A(x)=x(200-2x)$ ?
Answer: $x=50$ 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.
A farmer has $200$ m of fencing for three sides of a rectangle against a river; if depth is $x$ , what $x$ maximizes area?
Answer: $x=50$ 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.
A rectangle has area $A=x(18-x)$ for $0<x<18$ ; which $x$ maximizes $A$ ?
Answer: $x=9$ 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.
A lifeguard runs then swims; time is $T(x)=\frac{x}{6}+\frac{20-x}{2}$ for $0\le x\le 20$ ; which $x$ minimizes $T$ ?
Answer: $20$ This problem involves solving an optimization problem to minimize travel time. Given T(x) = x/6 + (20-x)/2, we find T'(x) = 1/6 - 1/2 = 1/6 - 3/6 = -2/6 = -1/3. Since T'(x) is constant and negative, T(x) is decreasing throughout [0,20], so the minimum occurs at x = 20. We can verify: T(0) = 0 + 10 = 10 hours, while T(20) = 20/6 + 0 = 10/3 ≈ 3.33 hours. The answer x = 0 might seem intuitive (swim the whole way), but running is faster (6 units/hour vs 2 units/hour). When optimizing linear functions, check if the function is monotonic—extrema occur at domain endpoints.
For $0<x<\frac{\pi}{2}$ , maximize $A(x)=\sin x\cos x$ ; which $x$ gives the maximum?
Answer: $x=\frac{\pi}{4}$ This problem requires solving optimization problems using calculus, a key skill in AP Calculus AB. To maximize A(x) = sin x cos x, compute the derivative A' = cos²x - sin²x and set it to zero, yielding x = π/4. Evaluating at x = π/4 gives the maximum, while endpoints approach zero. The second derivative confirms a maximum. A tempting distractor is x = π/3, but it yields a smaller value. 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.
For $x>0$ , the surface area of a closed cylinder with volume $100\pi$ is $S(x)=2\pi x^2+\dfrac{200\pi}{x}$ . Which $x$ minimizes $S(x)$ ?
Answer: $x=\sqrt[3]{50}$ This problem optimizes surface area for a cylinder with fixed volume. To minimize S(x) = 2πx² + 200π/x, we find S'(x) = 4πx - 200π/x². Setting S'(x) = 0 gives 4πx - 200π/x² = 0, so 4x³ = 200, which means x³ = 50, giving x = ∛50. We verify this is a minimum since S''(x) = 4π + 400π/x³ > 0 for all x > 0. The choice x = √50 might seem natural but confuses the cube root with square root. When optimizing surface area with volume constraints, the derivative often leads to solving a cubic equation.
For $0<x<\pi$ , maximize $f(x)=\sin x - \frac{x}{10}$ ; which $x$ gives the maximum?
Answer: $x=\arccos\left(\frac{1}{10}\right)$ 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.
A manufacturer's profit is $P(x)=-2x^2+80x-300$ for $0\le x\le 40$ . Which $x$ maximizes $P(x)$ ?
Answer: $x=20$ This profit optimization problem involves a quadratic function. To maximize P(x) = -2x² + 80x - 300, we find P'(x) = -4x + 80. Setting P'(x) = 0 gives -4x + 80 = 0, so x = 20. Since P''(x) = -4 < 0, this confirms x = 20 maximizes profit. The endpoint x = 40 might seem appealing as the upper bound, but P(40) = -2(1600) + 3200 - 300 = -300, which is actually a loss. For quadratic profit functions with negative leading coefficient, the maximum always occurs at the critical point, not the boundaries.
A ball is thrown upward with height $h(t)=-16t^2+64t+5$ (feet). At what $t$ is $h(t)$ maximized?
Answer: $2$ This problem requires solving an optimization problem to find when the ball reaches maximum height. Given h(t) = -16t² + 64t + 5, we find the maximum by taking the derivative h'(t) = -32t + 64. Setting h'(t) = 0 yields t = 2 seconds. We verify this is a maximum since h''(t) = -32 < 0, confirming the parabola opens downward. A common mistake is using the quadratic formula to find when h(t) = 0 (when the ball hits the ground) instead of finding when h'(t) = 0. For projectile motion problems, the maximum height always occurs when the velocity (first derivative) equals zero.
A right triangle has legs $x$ and $12-x$ for $0<x<12$ . Which $x$ maximizes its area $A(x)=\tfrac12 x(12-x)$ ?
Answer: $x=6$ 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.
A ladder of length $10$ ft reaches a wall; the base is $x$ ft from the wall, so height is $\sqrt{100-x^2}$ . Which $x$ maximizes the area of the right triangle formed?
Answer: $x=\frac{10}{\sqrt{2}}$ This problem requires solving optimization problems using calculus, a key skill in AP Calculus AB. To maximize the area A = (1/2) x √(100 - x²), compute the derivative and set it to zero, yielding x = 10/√2. Evaluating at x = 10/√2 gives the maximum area. The second derivative test confirms a maximum. A tempting distractor is x = 5, but it yields a smaller area. 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.
A company's profit is $P(x)=-2x^2+80x-300$ dollars for $0\le x\le 40$ . Which $x$ maximizes profit?
Answer: $20$ This problem requires solving an optimization problem to maximize profit. Given P(x) = -2x² + 80x - 300 for 0 ≤ x ≤ 40, we find the maximum by taking the derivative P'(x) = -4x + 80. Setting P'(x) = 0 yields x = 20, which lies within the given domain. We verify this is a maximum since P''(x) = -4 < 0, confirming the parabola opens downward. A common mistake is checking only the endpoints x = 0 or x = 40, missing the interior critical point. For optimization on a closed interval, always check both critical points and endpoints to find the global extremum.
A rectangle has one side on the $x$ -axis and top corners on $y=9-x^2$ ; if the right top corner is at $x$ , which $x$ maximizes area $A=2x(9-x^2)$ ?
Answer: $x=\sqrt{3}$ 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.
A rectangle has diagonal $10$ ; if one side is $x$ , the other is $\sqrt{100-x^2}$ . Which $x$ maximizes area $A=x\sqrt{100-x^2}$ ?
Answer: $x=\frac{10}{\sqrt{2}}$ This problem involves solving optimization problems by maximizing the area A(x) = x √(100 - x²) of a rectangle with diagonal 10. To find the maximum, compute the derivative A'(x) = (100 - 2x²)/√(100 - x²) and set it to zero, yielding x = 10/√2 as the critical point. Evaluating A at x = 10/√2 gives the maximum of 50. The interval is 0 < x < 10, and endpoints give A=0. A tempting distractor is x = 5, where A=5√75 ≈ 43.3, less than 50. 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.
A triangle has two sides of lengths $x$ and $12-x$ with included angle $\frac{\pi}{3}$ ; area is $A(x)=\frac{1}{2}x(12-x)\sin\frac{\pi}{3}$ . Which $x$ maximizes area?
Answer: $x=6$ This problem involves solving optimization problems by maximizing the area A(x) = (√3/4) x (12 - x) of a triangle. To find the maximum, compute the derivative A'(x) proportional to 12 - 2x and set it to zero, yielding x = 6 as the critical point. Evaluating A at x = 6 gives the maximum area. The interval is implicit, but the critical point determines the solution without endpoints. A tempting distractor is x = 0, where area is zero, but this is a minimum at 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.

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