Question-32

A rectangular, wire-fenced garden is to be constructed, one of whose sides is a wall. Find the maximum area that can be covered if the length of fencing wire available is \(\displaystyle L\).

• \(L^2\)

• \(\cfrac{L^2}{4}\)

• \(\cfrac{L^2}{8}\)

• \(\cfrac{L}{4}\)

Hint

Concave downards parabola

Answer

• \(L^2\)

• \(\cfrac{L^2}{4}\)

• \(\cfrac{L^2}{8}\)

• \(\cfrac{L}{4}\)

Solution

Let the dimensions of the garden be \(\displaystyle ( x,L-2x)\).

We have the following optimization problem:

\[ \begin{equation*} \underset{x}{\max} \ \ x( L-2x) \end{equation*} \]

Set \(\displaystyle f( x) =Lx-2x^{2}\). With \(\displaystyle f^{\prime }( x) =0\), we get \(\displaystyle x=\frac{L}{4}\). This corresponds to a maximum since \(\displaystyle f\) is a concave downwards parabola. Thus, the maximum area that can be enclosed is \(\displaystyle \frac{L}{4} \times \frac{L}{2} =\frac{L^{2}}{8}\).