Calculus Applications of differentiation

Optimization problems

Q: What is the objective function?

The quantity you are trying to make largest or smallest, such as area, product, or cost, written as a function of a single variable.

Q: How do you confirm a maximum rather than a minimum?

Use the second derivative test: the second derivative negative at the critical point means a maximum, positive means a minimum.

Q: What rectangle of fixed perimeter has the most area?

A square. For a perimeter of 40 feet the best rectangle is 10 by 10 feet, giving 100 square feet.

20 practice questions 0 video lessons Theory + worked examples

Optimization Problems

California Calculus • Standard 11.0 • Applications of Differentiation

Optimization Problems is a topic in Applications of Differentiation in the California Calculus Standards. It is aligned to Standard 11.0, which requires students to use differentiation to solve optimization (maximum-minimum) problems in a variety of applied contexts.

Optimization uses derivatives to find the largest or smallest value in an applied problem. You write the quantity to optimize as a function of one variable, set its derivative to zero, and confirm a maximum or minimum.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Theory

Optimization uses derivatives to find the largest or smallest value in an applied problem — the biggest area, the cheapest box, the best product. Write the quantity to optimize as a function of one variable, set its derivative to zero, and confirm whether it is a max or a min.

An optimization problem asks for the maximum or minimum of some quantity subject to a constraint. The plan:

Write the objective (what you optimize) as a function of one variable, using the constraint to eliminate the other.
Find the critical point by setting the derivative to \(0\).
Confirm max versus min, then report the quantity actually asked.

Key idea: answer the question asked. Sometimes it wants a dimension (\(x=10\) ft), sometimes the optimized value (area \(=100\) sq ft) — read carefully.

The objective \(A=x(20-x)\) comes from the constraint.

The area function peaks at the optimal \(x\).

The optimization routine:

\[\text{optimize } Q=f(x)\ \Rightarrow\ f'(x)=0\ \Rightarrow\ \text{check } f''\]

set the derivative of the objective to zero, then check the second derivative

Classic result: for a fixed perimeter, the rectangle of greatest area is a square.

How to solve an optimization problem

Name the variables and write the constraint.
Write the objective as a function of one variable.
Differentiate and solve \(f'(x)=0\).
Confirm max or min with \(f''\), and report the requested quantity with units.

Example 1 — Maximum area

A rectangle has perimeter \(40\) ft. What dimensions give the greatest area?

Solution

With width \(x\), the height is \(20-x\), so the objective is \(A=x(20-x)=20x-x^2\).

\(A'(x)\)	\(=\)	\(20-2x=0\)
\(x\)	\(=\)	\(10\)

\(A''=-2<0\) confirms a max. The rectangle is \(10\times 10\) ft (a square), area \(100\) sq ft.

maximum area is a 10 by 10 square, 100 square feet

Example 2 — Maximum product

Two positive numbers add to \(12\). Maximize their product.

Solution

Let the numbers be \(x\) and \(12-x\); the product is \(P=x(12-x)=12x-x^2\).

\(P'(x)\)	\(=\)	\(12-2x=0\)
\(x\)	\(=\)	\(6\)

The numbers are \(6\) and \(6\); the maximum product is \(36\).

maximum product is 36 with both numbers 6

Example 3 — Fencing against a wall

\(100\) ft of fence encloses a rectangular yard against a wall (three sides). Maximize the area.

Solution

With the two ends \(x\) and the far side \(100-2x\), \(A=x(100-2x)=100x-2x^2\).

\(A'(x)\)	\(=\)	\(100-4x=0\)
\(x\)	\(=\)	\(25\)

Dimensions \(25\times 50\) ft; maximum area \(1250\) sq ft.

maximum area 1250 square feet, 25 by 50 feet

Example 4 — Confirm max vs min

For \(A=x(20-x)\), how do you know \(x=10\) is a maximum?

Solution

Use the second derivative at the critical point.

\(A''(x)\)	\(=\)	\(-2\)
\(A''(10)\)	\(=\)	\(-2<0\)

\(A''<0\), so the critical point is a maximum.

second derivative negative confirms a maximum

Common pitfalls

Reduce to one variable first. Use the constraint to eliminate the extra variable before differentiating.

Confirm max vs min. A critical point is not automatically the maximum — check \(f''\) or the endpoints.

Answer the actual question. If it asks for the area, do not stop at the dimension; substitute back to get the value.

Frequently asked questions

How do you solve an optimization problem in calculus?

Write the quantity to optimize as a function of one variable using the constraint, set its derivative to zero, confirm max or min with the second derivative, and report the requested value.

What is the objective function?

The quantity you are trying to make largest or smallest — area, product, cost — written as a function of a single variable.

How do you confirm a maximum rather than a minimum?

Use the second-derivative test: \(f''<0\) at the critical point means a maximum, \(f''>0\) means a minimum.

What rectangle of fixed perimeter has the most area?

A square. For a perimeter of \(40\) ft the best rectangle is \(10\times 10\) ft, giving \(100\) sq ft.