Optimisation problems

General approach

To solve an optimisation problem using differentiation:

1. Model the situation using a single-variable function.
2. Determine the domain of definition, taking into account physical constraints (positive lengths, etc.).
3. Differentiate the function and examine the sign of the derivative.
4. Determine the nature of the extremum (maximum or minimum) and calculate its value.

Example

A box without a lid is made from a square sheet measuring 12 cm on each side, by cutting out squares of side length x at each corner and then folding the edges. The volume is:

V(x) = x * (12 - 2x)², where 0 < x < 6

We differentiate: V'(x) = (12 - 2x)² - 4x(12 - 2x) = (12 - 2x) * [(12 - 2x) - 4x] = (12 - 2x)(12 - 6x)

V'(x) = 0 for x = 6 (excluded from the domain) or x = 2.

• For 0 < x < 2: V'(x) > 0, V increases.
• For 2 < x < 6: V'(x) < 0, V decreases.

Therefore, V has a maximum at x = 2, with V(2) = 2 * 8² = 128 cm³.

Pitfall

The domain of investigation must always be restricted to physically possible values. Here, x = 6 does indeed set V'(x) to zero but corresponds to a volume of zero (and is therefore irrelevant), and values of x outside ]0; 6[ are meaningless for this cutting problem.