Practice Problem Set: Optimization

How to Approach Optimization Problems

Optimization problems ask you to find the largest or smallest possible value of some quantity. The calculus is usually only one part of the problem. The harder part is often translating the verbal description into a mathematical function that can actually be differentiated.

A strong AP Calculus solution should make the objective function explicit, identify the relevant domain, find the critical point or points, and explain why the resulting value is the desired maximum or minimum.

The Optimization Checklist

Identify what is being optimized. Is the problem asking for maximum area, minimum cost, shortest distance, maximum volume, or something similar?
Define variables. Give the dimensions or quantities mathematical names.
Write the constraint. Use the information in the problem to connect the variables.
Write the objective function. Express the quantity being optimized in terms of one variable whenever possible.
Determine the domain. Physical dimensions and other restrictions matter.
Differentiate and find critical points. Solve $f'(x)=0$ and check where $f'(x)$ is undefined if relevant.
Test candidates. Use the First Derivative Test, Second Derivative Test, or absolute-extrema reasoning.
Answer the actual question. State the requested dimension, quantity, or value with appropriate units.

The Core Optimization Idea

Suppose the quantity you want to optimize is represented by $Q(x)$ on a closed interval $[a,b]$. An absolute maximum or minimum can occur at a critical point in the interior or at an endpoint.

\[ Q'(x)=0 \]
Critical points occur where the derivative is zero or does not exist.

In an optimization problem, solving $Q'(x)=0$ is not automatically enough. You still have to determine whether the critical point produces the desired maximum or minimum and whether it is physically meaningful.

Worked Examples

Example 1: Maximum Area with a Fixed Perimeter

A farmer has 200 meters of fencing and wants to build a rectangular enclosure. What dimensions maximize the area?

Step 1: Let the rectangle have length $x$ and width $y$.

Step 2: The fencing constraint is \[ 2x+2y=200. \] Therefore, \[ y=100-x. \]

Step 3: The area is \[ A=xy. \] Substitute the constraint: \[ A(x)=x(100-x)=100x-x^2. \]

Step 4: Differentiate: \[ A'(x)=100-2x. \] Set the derivative equal to zero: \[ 100-2x=0 \] so \[ x=50. \]

Step 5: The corresponding width is \[ y=100-50=50. \]

Conclusion: The maximum-area rectangle is a $50$-meter by $50$-meter square, with maximum area \[ \boxed{2500\text{ m}^2}. \]

Example 2: Minimum Material for an Open-Top Box

An open-top box must have a square base and a volume of $32\text{ ft}^3$. Find the dimensions that minimize the amount of material used.

Step 1: Let the side length of the square base be $x$, and let the height be $h$.

Step 2: The volume constraint is \[ x^2h=32. \] Thus, \[ h=\frac{32}{x^2}. \]

Step 3: Because the box has no top, the surface area is \[ S=x^2+4xh. \] Substitute $h$: \[ S(x)=x^2+4x\left(\frac{32}{x^2}\right) =x^2+\frac{128}{x}. \]

Step 4: Differentiate: \[ S'(x)=2x-\frac{128}{x^2}. \] Set equal to zero: \[ 2x-\frac{128}{x^2}=0. \] Multiplying by $x^2$, \[ 2x^3=128, \] so \[ x^3=64 \] and therefore \[ x=4. \]

Step 5: Find the height: \[ h=\frac{32}{4^2}=2. \]

Conclusion: The dimensions that minimize material are a square base measuring $4\text{ ft}$ by $4\text{ ft}$ with height $2\text{ ft}$.

Example 3: Minimum Distance to a Point

Find the point on the parabola $y=x^2$ that is closest to the point $(0,3)$.

Step 1: A point on the parabola has coordinates \[ (x,x^2). \]

Step 2: The distance to $(0,3)$ is \[ D=\sqrt{x^2+(x^2-3)^2}. \]

Step 3: Because the square-root function is increasing, minimizing $D$ is equivalent to minimizing $D^2$: \[ D^2=x^2+(x^2-3)^2. \] Expand: \[ D^2=x^4-5x^2+9. \]

Step 4: Differentiate: \[ \frac{d}{dx}(D^2)=4x^3-10x. \] Set equal to zero: \[ 4x^3-10x=0. \] Factor: \[ 2x(2x^2-5)=0. \] Thus, \[ x=0,\qquad x=\pm\sqrt{\frac52}. \]

Step 5: Compare the corresponding values of $D^2$. At \[ x=\pm\sqrt{\frac52}, \] the minimum occurs.

Conclusion: The closest points are \[ \boxed{\left(\sqrt{\frac52},\frac52\right)} \quad\text{and}\quad \boxed{\left(-\sqrt{\frac52},\frac52\right)}. \]

Example 4: Cost Minimization

A company wants to build a rectangular storage area with an area of $600\text{ ft}^2$. The two longer sides cost \$8 per foot to construct, while the two shorter sides cost \$12 per foot. Find the dimensions that minimize the total construction cost.

Step 1: Let the length be $x$ and width be $y$. The area constraint is \[ xy=600. \] Therefore, \[ y=\frac{600}{x}. \]

Step 2: The two longer sides have total cost $16x$, while the two shorter sides have total cost $24y$. Thus, \[ C=16x+24y. \] Substitute the constraint: \[ C(x)=16x+\frac{14400}{x}. \]

Step 3: Differentiate: \[ C'(x)=16-\frac{14400}{x^2}. \] Set equal to zero: \[ 16-\frac{14400}{x^2}=0. \] Therefore, \[ 16x^2=14400 \] and \[ x^2=900. \] Since a length must be positive, \[ x=30. \]

Step 4: Find the width: \[ y=\frac{600}{30}=20. \]

Conclusion: The cost is minimized when the storage area is \[ \boxed{30\text{ ft}\times20\text{ ft}}. \]

Common Optimization Mistakes

Optimizing the wrong quantity

The derivative must be taken of the quantity the question actually asks you to maximize or minimize. If the problem asks for maximum area, your objective function should be area—not a constraint such as perimeter.

Keeping two variables in the objective function

The constraint is usually there so that you can rewrite the objective function using one variable. If your objective still contains both $x$ and $y$, look for a way to substitute using the constraint.

Ignoring the domain

A critical point is only useful if it makes sense in the original problem. A negative length, a dimension of zero, or a value outside the permitted interval is not a valid physical solution.

Stopping after finding $f'(x)=0$

Finding a critical point does not automatically prove that it is the requested maximum or minimum. Use the Second Derivative Test, First Derivative Test, or compare all relevant candidates.

Giving only the numerical answer

AP free-response questions often award points for defining variables, writing the objective function, showing the derivative equation, and interpreting the answer. A correct number without the supporting work may not receive full credit.

Practice Problems

Question 1 — Maximum Area

A rectangular garden is built against a straight river, so no fencing is needed along the river. A farmer has 240 meters of fencing for the other three sides. What dimensions maximize the area of the garden?

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Let $x$ be the width perpendicular to the river and $y$ the side parallel to the river. The constraint is \[ 2x+y=240, \] so \[ y=240-2x. \] The area is \[ A(x)=x(240-2x)=240x-2x^2. \] Thus \[ A'(x)=240-4x. \] Setting $A'(x)=0$ gives $x=60$. Then \[ y=240-120=120. \] Therefore the maximum-area garden is \[ \boxed{60\text{ m}\times120\text{ m}}. \] The maximum area is \[ \boxed{7200\text{ m}^2}. \]

Question 2 — Minimum Surface Area

A closed cylindrical container must hold $500\pi\text{ cm}^3$. Find the radius and height that minimize the surface area.

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The volume constraint is \[ \pi r^2h=500\pi, \] so \[ h=\frac{500}{r^2}. \] Surface area is \[ S=2\pi r^2+2\pi rh. \] Therefore \[ S(r)=2\pi r^2+\frac{1000\pi}{r}. \] Differentiate: \[ S'(r)=4\pi r-\frac{1000\pi}{r^2}. \] Set equal to zero: \[ 4\pi r=\frac{1000\pi}{r^2}. \] Hence \[ 4r^3=1000, \qquad r^3=250. \] Thus \[ \boxed{r=\sqrt[3]{250}}. \] The corresponding height is \[ h=\frac{500}{r^2}=2\sqrt[3]{250}. \] Therefore \[ \boxed{r=\sqrt[3]{250}\text{ cm},\qquad h=2\sqrt[3]{250}\text{ cm}}. \]

Question 3 — Minimum Distance

Find the point on the curve $y=x^2+1$ that is closest to the point $(2,0)$.

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A point on the curve is \[ (x,x^2+1). \] Minimize the squared distance: \[ D^2=(x-2)^2+(x^2+1)^2. \] Expanding, \[ D^2=x^4+3x^2-4x+5. \] Differentiate: \[ (D^2)'=4x^3+6x-4. \] Setting equal to zero gives \[ 2x^3+3x-2=0. \] The relevant real solution is approximately \[ x\approx0.596. \] Therefore \[ y=x^2+1\approx1.355. \] The closest point is approximately \[ \boxed{(0.596,\;1.355)}. \]

Question 4 — Optimization with a Parameter

A rectangle has one side along the $x$-axis and its upper corners lie on the parabola $y=12-x^2$. Find the dimensions of the rectangle with maximum area.

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By symmetry, let the upper-right corner be \[ (x,12-x^2). \] The rectangle has width $2x$ and height $12-x^2$. Thus \[ A(x)=2x(12-x^2)=24x-2x^3. \] Differentiate: \[ A'(x)=24-6x^2. \] Set equal to zero: \[ 24-6x^2=0, \] so \[ x^2=4 \] and therefore \[ x=2. \] The height is \[ 12-2^2=8. \] The width is \[ 2(2)=4. \] Therefore the maximum-area rectangle has dimensions \[ \boxed{4\times8} \] and maximum area \[ \boxed{32}. \]

Question 5 — Cost Optimization

A company needs a rectangular enclosure with area $1000\text{ ft}^2$. The cost of fencing one pair of opposite sides is \$5 per foot, while the other pair costs \$10 per foot. Find the dimensions that minimize the total fencing cost.

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Let $x$ be the sides costing \$5 per foot and $y$ the sides costing \$10 per foot. The area constraint is \[ xy=1000, \] so \[ y=\frac{1000}{x}. \] The cost is \[ C=10x+20y. \] Therefore \[ C(x)=10x+\frac{20000}{x}. \] Differentiate: \[ C'(x)=10-\frac{20000}{x^2}. \] Set equal to zero: \[ 10=\frac{20000}{x^2}. \] Thus \[ x^2=2000, \qquad x=20\sqrt5. \] Then \[ y=\frac{1000}{20\sqrt5}=10\sqrt5. \] Therefore the dimensions are \[ \boxed{20\sqrt5\text{ ft}\times10\sqrt5\text{ ft}}. \]

AP Exam Strategy

On an AP Calculus free-response question, optimization is rarely just a derivative computation. The scoring points are often distributed across the setup, derivative, critical-point equation, and interpretation.

A Reliable Free-Response Structure

Define the variable clearly.
Write the constraint equation.
Use the constraint to obtain a one-variable objective function.
State the relevant domain.
Differentiate.
Set the derivative equal to zero and solve for the critical point.
Explain why the critical point gives the requested maximum or minimum.
State the final answer in context and include units when appropriate.

A particularly useful shortcut is to optimize a simpler equivalent quantity when possible. For example, when minimizing a distance \[ D=\sqrt{F(x)}, \] you can minimize $F(x)=D^2$ instead because the square-root function is increasing.

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