Optimisation problems

Optimisation means finding the best: the largest area, the least cost, the smallest material. The method:

• Name the quantity to make largest or smallest, and write a formula for it.
• Use the given condition to remove extra variables, so the formula has one variable only.
• Differentiate, set the derivative to 0, solve.
• Check it is really a maximum or minimum (second derivative), and that the answer makes sense: no negative or impossible lengths.

Example: a rectangle has perimeter 40 m. Find its largest area.

• Sides x and y. Area A = xy.
• Condition: 2x + 2y = 40, so y = 20 – x, and A = x(20 – x) = 20x – x².
• A′ = 20 – 2x = 0 gives x = 10.
• A″ = –2 < 0, so this is a maximum. y = 10, so the best rectangle is a square, area 100 m².

Drawing a diagram first helps a lot, especially for boxes, cones and shapes inside circles.

Box from a sheet: cutting a square of side x from each corner of a sheet and folding up the sides gives a box of height x. Write the volume in x, then maximise. x must be positive and less than half the sheet's side, so reject any root outside that range.

For shapes inside a circle or sphere, a trick helps: maximise the square of the quantity (or its volume without π). It has its maximum at the same point, and the algebra is easier.