Teach Me Maths

Maximum and Minimum Problems

How to form an expression from a constraint, use calculus to optimise it, and check the answer against the allowed domain.

What an optimisation problem asks

An optimisation problem asks for the greatest or least possible area, volume or other quantity. Usually there is a constraint, such as a fixed perimeter or volume. Use that constraint to write the quantity in terms of one variable.

  • Choose a variable and state its allowed values.
  • Use the constraint to eliminate other variables.
  • Differentiate the quantity being optimised and solve for zero gradient.
  • Reject values outside the domain and justify maximum or minimum.
  • Substitute into the original expression and answer with the correct units.

If the question says use calculus, show the derivative and the equation you solve. Testing a few values or quoting a known shape is not a calculus argument.

Worked example 1

A rectangle has perimeter 36 cm. Use calculus to find its maximum area.

  1. Let the width be x cm and the length be l cm. Then 2x+2l=36, so l=18-x and 0<x<18
  2. The area is A=x(18-x)=18x-x^2. Hence \frac{dA}{dx}=18-2x

  3. Your turn. Solve 18-2x=0

    x=9, so l=9

  4. \frac{d^2A}{dx^2}=-2<0, so this gives a maximum. The area approaches 0 at either end of the allowed interval.
  5. Answer: the maximum area is 81\text{ cm}^2, for a 9 cm by 9 cm square.

Explore a fixed perimeter

Try it: compare widths 6 and 12. They give the same area because the rectangle’s dimensions swap. Watch the gradient change from positive to negative at x=9

Worked example 2

A rectangular enclosure uses 48 m of fencing for three sides. A straight wall forms the fourth side. Find the dimensions that maximise the enclosed area.

  1. Let each side perpendicular to the wall be x m, and the side parallel to it be y m. Only three sides need fencing: 2x+y=48

  2. Your turn. Write y in terms of x

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    y=48-2x, with 0<x<24

  3. A=x(48-2x)=48x-2x^2, so \frac{dA}{dx}=48-4x. Set this to zero: x=12
  4. \frac{d^2A}{dx^2}=-4<0. At x=12, the other dimension is 48-24=24
  5. Answer: 12 m perpendicular to the wall and 24 m parallel to it, giving maximum area 288\text{ m}^2

Use the physical domain

Lengths must be positive. If a cut of size x removes material from both ends of a side of length L, the remaining length is L-2x. A root that makes a dimension zero or negative is not a possible object.

Worked example 3

Squares of side x cm are cut from all four corners of a 32 cm by 20 cm sheet. The sides are folded up to make an open box. Find the maximum volume.

  1. The box has height x and base dimensions 32-2x and 20-2x. The domain is 0<x<10
  2. V=x(32-2x)(20-2x)=640x-104x^2+4x^3
  3. \frac{dV}{dx}=640-208x+12x^2=4(3x-40)(x-4)

  4. Your turn. The roots are 4 and \frac{40}{3}. Which root is in the domain 0<x<10?

    Only x=4 is valid; \frac{40}{3}>10

  5. \frac{d^2V}{dx^2}=24x-208, which is -112 at x=4. This is a maximum; the volume tends to zero at both ends of the domain.
  6. Answer: V=4(24)(12)=1152\text{ cm}^3. The cut size is 4 cm.

Worked example 4

For x>0, use calculus to find the minimum value of F=3x^2+\frac{48}{x}.

  1. Write F=3x^2+48x^{-1}. Then \frac{dF}{dx}=6x-48x^{-2}
  2. Set the derivative to zero and multiply by x^2: 6x^3-48=0, so x^3=8

  3. Your turn. Solve x^3=8

    x=2

  4. \frac{d^2F}{dx^2}=6+\frac{96}{x^3}>0 for every x>0. The derivative changes from negative to positive at x=2, so this is the minimum over the domain.
  5. Answer: the minimum value is 3(2)^2+\frac{48}{2}=36, when x=2

A fixed volume and a variable surface area

For a tank or box, distinguish the fixed quantity from the quantity you are minimising. Use the volume constraint to eliminate the height, then differentiate the surface area. Count only the faces the question includes.

Worked example 5

An open-topped tank has a square base of side x cm and volume 500\text{ cm}^3. Find the dimensions that minimise its surface area.

  1. If its height is h, then x^2h=500, so h=\frac{500}{x^2} and x>0
  2. There is one base and four side faces: S=x^2+4xh=x^2+\frac{2000}{x}
  3. \frac{dS}{dx}=2x-\frac{2000}{x^2}=0 gives 2x^3=2000, so x=10

  4. Your turn. Find h=\frac{500}{x^2} when x=10

    h=\frac{500}{100}=5 cm.

  5. \frac{d^2S}{dx^2}=2+\frac{4000}{x^3}>0 for x>0. The area grows without bound as x approaches 0 or increases without bound.
  6. Answer: a 10 cm by 10 cm base and height 5 cm. The minimum surface area is 100+200=300\text{ cm}^2

Check endpoints when they are included

On a closed interval, compare all stationary values in the interval with the values at both endpoints. A largest or smallest value can occur at an endpoint even though its gradient is not zero.

Worked example 6

Find the greatest value of y=x^3-3x for -2\le x\le3.

  1. \frac{dy}{dx}=3x^2-3=0 gives x=-1 or x=1, both in the interval.
  2. The stationary values are y=2 at x=-1 and y=-2 at x=1
  3. At the left endpoint, y=(-2)^3-3(-2)=-2

  4. Your turn. Find y at the other endpoint, x=3

    y=3^3-3(3)=18

  5. Answer: the greatest value is 18, at x=3. The local maximum value 2 is not the greatest value on this interval.

Common mistakes

  • Differentiating with two variables still present. Use the constraint to write the target quantity in one variable.
  • Optimising the fixed quantity. Write an expression for the quantity the question asks you to maximise or minimise.
  • Removing material from one end only. When cutting a box net, subtract the cut size from both ends of each affected side.
  • Keeping an impossible dimension. Reject stationary values outside the physical domain.
  • Giving only the variable value. Use it to find the requested dimensions, area or volume, with units.
  • Assuming a stationary point gives the required optimum. Justify its type and compare any included endpoints.

Now try it: Maximum and Minimum Problems practice questions

More on this topic: Maximum and Minimum Problems worksheet with full solutions