Skip to main content
Mathovia

Calculus / Calculus 1 · Applications of Derivatives

Optimization Problems in Calculus

Optimization means finding the largest or smallest value of a quantity: the most area, the least material, the shortest time. Write the quantity as a function of one variable, find its critical points, and check which one gives the absolute maximum or minimum on the allowed interval.

Practice Problems
Written by
MPhil in Mathematics, Founder of Mathovia
Published

Mathematics verification: Mathematical formulas, calculations, worked examples and solutions are reviewed for accuracy before publication. How we check our work

Chapter 4 · Lesson 10 of 10Calculus 1 · Applications of Derivatives
  1. 1.Related Rates
  2. 2.Linear Approximation
  3. 3.Critical Points
  4. 4.Extreme Value Theorem
  5. 5.Rolle's Theorem
  6. 6.Mean Value Theorem
  7. 7.First Derivative Test
  8. 8.Second Derivative Test
  9. 9.Concavity and Inflection Points
  10. 10.Optimization Problems in Calculus
See all calculus chapters →

How much area can 2,400 feet of fence enclose? What’s the cheapest shape for a soda can? Where should you land a boat to get to town fastest? These are optimization problems: find the largest or smallest value of some quantity. Calculus solves all of them the same way. Write the quantity as a function of one variable, find the critical points, and check which one gives the absolute maximum or minimum.

Maximize A of x equals x times 2400 minus 2x: A prime of x equals 2400 minus 4x equals 0, so x equals 600
The idea in one line: set the derivative of the quantity to zero

The Steps for Solving Optimization Problems

  1. Read and draw. Sketch the situation and label every quantity that can change.
  2. Name the goal. Write a formula for the quantity you want to maximize or minimize (the objective function).
  3. Use the constraint. Find the equation that links the variables, solve it for one of them, and substitute. The objective is now a function of one variable.
  4. Find the domain. Decide which values of the variable make physical sense.
  5. Differentiate and solve for the critical points in the domain.
  6. Confirm it’s the absolute max or min (see the next section).
  7. Answer the question. Reread it. Does it ask for the dimensions, the maximum value, or both? Include units.

Most of the work is in steps 1 to 4. Once you have the right function, the calculus is usually short.

Making Sure It’s the Absolute Max or Min

A critical point is only a candidate. You need a reason it gives the absolute maximum or minimum, not just a local one. Use whichever fits:

  • Closed interval method. If the domain is a closed interval , evaluate at the critical points and the endpoints and compare. This is the method from the extreme value theorem lesson.
  • Only one critical point. If is continuous on an interval and has just one critical point there, and the first derivative test shows it’s a local max, it’s the absolute max. (The same goes for a min.)
  • One concavity. If on the whole domain, any critical point is the absolute max; if everywhere, it’s the absolute min. This is the second derivative test applied to the whole interval.

Example 1: Fencing Along a River

A farmer has 2,400 ft of fencing to enclose a rectangular field along a straight river. No fence is needed along the river. What dimensions give the largest area?

Left: a rectangular field along a river, fenced on three sides. The two sides perpendicular to the river have length x and the side parallel to it has length y, with 2x plus y equal to 2400 feet. Right: the graph of A of x equals x times 2400 minus 2x, a downward parabola from x equals 0 to 1200, with its maximum at x equals 600
The constraint turns the area into a function of one variable

Goal: maximize , where is the side perpendicular to the river and is the side parallel to it.

Constraint: the fence covers two sides of length and one of length , so , and .

One variable:

Critical point: at .

Check: on the closed interval , , and . The maximum is at .

Answer: the field should be 600 ft by 1,200 ft (the long side along the river), enclosing 720,000 ft².

Example 2: The Open Box

A box with an open top is made from a 12-inch square of cardboard by cutting an -by- square from each corner and folding up the sides. What value of gives the largest volume?

Left: a 12 inch square sheet with an x by x square cut from each corner and dashed fold lines, leaving a base of side 12 minus 2x. Right: the folded open box, with height x and base 12 minus 2x by 12 minus 2x, and volume V of x equals x times 12 minus 2x squared, for x between 0 and 6
Height x, square base of side 12 − 2x

After folding, the box has height and a square base of side . So

By the product rule and the chain rule,

Factor out :

at (an endpoint) and . Compare: , , .

Answer: cut 2-inch squares. The box is 8 in by 8 in by 2 in, with volume 128 in³.

Example 3: The Can That Uses the Least Material

A cylindrical can must hold 1,000 cm³. What radius makes the surface area (top, bottom and side) as small as possible?

Goal: minimize (two circles plus the side).

Constraint: , so . Substituting,

Critical point:

Check: for every , so is concave up on its whole domain and this critical point is the absolute minimum.

Then cm.

Answer: cm and cm. The cheapest can is as tall as it is wide.

Example 4: The Closest Point on a Curve

Find the point on the curve closest to the point .

A point on the curve is . Its distance to is . The square root makes the derivative messy, so minimize instead. It has its minimum at the same .

everywhere, so this is the absolute minimum on the domain .

Answer: the closest point is , at distance .

Example 5: The Fastest Route

You’re in a boat 3 km from the nearest point on a straight shore. A town is 8 km down the shore from that point. You can row at 3 km/h and walk at 5 km/h. Where should you land to reach the town as fast as possible?

A boat 3 kilometers offshore. The rower heads diagonally to a landing point x kilometers along the shore, rowing at 3 kilometers per hour, then walks the remaining 8 minus x kilometers to the town at 5 kilometers per hour
Row diagonally to the shore, then walk the rest

Land km down the shore. The rowing distance is by the Pythagorean theorem, and the walk is . Time is distance over speed:

Squaring both sides, , so and .

Check the closed interval :

h
h
h

Answer: land 2.25 km down the shore. The trip takes 2.4 hours, 12 minutes faster than rowing straight to shore.

Example 6: The Largest Rectangle Under a Parabola

A rectangle has its base on the x-axis and its top two corners on the parabola . Find its largest possible area.

By symmetry, put the corners at . The width is and the height is :

at . The endpoints give , and .

Answer: the largest area is 32, for a rectangle 4 wide and 8 tall.

Justifying Your Answer on the AP Calculus Exam

AP Calculus AB and BC free-response questions usually award a point for justifying that your answer is an absolute maximum or minimum. Finding isn’t enough. Write one of these:

  • Candidates test: “The absolute maximum on occurs at a critical point or an endpoint. , , , so the maximum is .” Show the table.
  • Only critical point: ” changes from positive to negative at , the only critical point on the interval, so has its absolute maximum at .”
  • Concavity: ” for all in the domain, so the critical point is the absolute maximum.”

Use the function’s name in the sentence, and make it clear which test you’re using.

Common Mistakes

  • Optimizing a function of two variables. Use the constraint to get down to one variable before differentiating.
  • Ignoring the domain. In Example 2, makes but it’s an endpoint with zero volume.
  • Skipping the check. A critical point can be a minimum when you wanted a maximum.
  • Answering the wrong question. If the problem asks for the dimensions, give both of them, not just .
  • Forgetting units. Area is in square units and volume in cubic units.

Practice Problems

Try each one before opening the answer. The practice set has 12 more with instant checking.

Problem 1. Find two positive numbers whose product is 100 and whose sum is as small as possible.

Show answer

With , the sum is for . at , and , so it’s the absolute minimum.

Answer: 10 and 10, with sum 20

Problem 2. A rectangular plot of 1,800 m² is fenced on three sides; the fourth side is a river. What is the least fencing needed?

Show answer

Let be the sides perpendicular to the river, so the parallel side is . The fencing is . at , and for .

Answer: 120 m of fence, for a plot 30 m by 60 m

Problem 3. An open box is made from a 10-by-16-inch sheet by cutting -by- squares from the corners. Find the largest volume.

Show answer

. Only is in the domain (). and .

Answer: 144 in³, with 2-inch cuts

Problem 4. Find the points on closest to .

Show answer

Minimize . Then , which is 0 at and . and .

Answer: , at distance

Problem 5. A theater sells 500 tickets at $20. For each $1 drop in price, it sells 50 more. What price maximizes revenue?

Show answer

After one-dollar drops, . at , and .

Answer: $15 a ticket, for revenue

Problem 6. Find the largest rectangle that fits in a semicircle of radius 5, with its base on the diameter.

Show answer

With top corners at , on .

The endpoints give 0, and .

Answer: area 25, for a rectangle wide and tall

Optimization brings together everything in this chapter: critical points, the extreme value theorem and the derivative tests. Next, curve sketching uses the same tools to draw an accurate graph from a formula.

Frequently Asked Questions

What is optimization in calculus?+

Optimization is using derivatives to find the maximum or minimum value of a quantity in a real situation, such as the largest area a fence can enclose or the cheapest can that holds a given volume. You write the quantity as a function, find where its derivative is zero, and confirm which point gives the absolute max or min.

What are the steps for solving optimization problems?+

Draw a picture and name the variables. Write the function to maximize or minimize. Use the constraint to rewrite it in one variable and find its domain. Differentiate and solve . Check that the critical point gives the absolute max or min, using the closed interval method or the first or second derivative test. Then answer the question that was asked, with units.

How do I know if I found a maximum or a minimum?+

On a closed interval, compare the function's values at the critical points and endpoints; the largest is the max and the smallest is the min. On an open interval with only one critical point, use the first or second derivative test: if it's a local max and the only critical point, it's the absolute max. If has one sign on the whole domain, the critical point is the absolute extremum.

How do you justify an optimization answer on the AP Calculus exam?+

State a reason that the critical point is an absolute extremum, not just a local one. Common justifications: changes from positive to negative at the only critical point, so it's the absolute maximum; or a table of values at the critical points and endpoints shows it's the largest. Saying 'f′ = 0' alone doesn't earn the point.

What is the constraint in an optimization problem?+

The constraint is the condition that ties the variables together, like a fixed amount of fencing or a fixed volume. It's an equation you use to eliminate a variable so the quantity you're optimizing depends on only one variable.

Why minimize the distance squared instead of the distance?+

Distance has a square root, which makes the derivative messy. Since the square root is increasing, the distance and its square are smallest at the same point. Minimize , then take the square root at the end.

Related lessons

Extreme Value TheoremSecond Derivative TestFirst Derivative Test