Optimisation

Level 1 - Calculus topic page in Applications Of Derivatives.

Principle

Optimisation finds the largest or smallest value of a quantity under stated conditions. Calculus identifies candidates using derivatives, then the context decides which candidate is relevant.

Notation

\(Q(x)\)

quantity being optimised

\(Q'(x)\)

derivative used to find critical points

\(D\)

allowed domain for the variable

The Core Method

Write the target quantity as a function of one variable, restrict the domain, solve \(Q'(x)=0\) or find endpoints, then compare values.

Critical point condition

\[Q'(x)=0\]

For closed intervals, endpoints must be checked. For open or physical domains, use sign changes, second derivatives, or context to justify the optimum.

Worked Cases

Question

A rectangle has perimeter \(20\). Find the dimensions that maximise its area.

Answer

Let the sides be \(x\) and \(y\). The perimeter condition gives \(2x+2y=20\), so \(y=10-x\). Area is \(A(x)=x(10-x)=10x-x^2\). Differentiate: \(A'(x)=10-2x\). Set \(A'(x)=0\): \(x=5\), and then \(y=5\). Since \(A\) is a downward-opening quadratic, this is the maximum.

Examples

Question

Find the minimum of \(f(x)=x^2-4x+7\) on \([0,5]\).

Answer

Differentiate: \(f'(x)=2x-4\). Set \(f'(x)=0\), giving \(x=2\). Check \(x=0,2,5\): \(f(0)=7\), \(f(2)=3\), and \(f(5)=12\). The minimum is \(3\) at \(x=2\).

Curve Sketching

Related Rates