Maximum-area pen against a wall

Statement

A farmer has $60\ \text{m}$ of fencing to enclose a rectangular area against a wall (the wall forms the fourth side and needs no fence). Find the dimensions that maximise the area and the value of that area.

Solution

Let $x$ be the two sides perpendicular to the wall and $60-2x$ the parallel side. The area is $A(x)=x(60-2x)=60x-2x^2.$ $A'(x)=60-4x=0\Rightarrow x=15\ \text{m}$; since $A''(x)=-4<0$, it is a maximum.

The parallel side is $60-30=30\ \text{m}$ and the maximum area is $A(15)=15\cdot 30=450\ \text{m}^2$.

$\boxed{x=15\ \text{m},\ \text{side }30\ \text{m},\ A_{\max}=450\ \text{m}^2}$