# A farmer has 240 meters of fencing material to enclose his rectangular plot of land. However, it is bound on one side by a river, and on the opposite side, half the fencing is purchased and supplied by the farmer that owns the lot. See below?

## What is the maximum area of the farmer can make with this amount of fencing material and what will be the dimensions?

Sep 24, 2016

Let $x$ be the length and $y$ the width.

The perimeter is given by:

$P = x + \frac{x}{2} + 2 y$

$240 = x + \frac{x}{2} + 2 y$

$240 = \frac{2 x + x + 4 y}{2}$

$480 = 3 x + 4 y$

$- \frac{3}{4} x + 120 = y$

The area is given by:

A = xy

$A = \left(- \frac{3}{4} x + 120\right) \left(x\right)$

$A = - \frac{3}{4} {x}^{2} + 120 x$

The maximum area will be the vertex:

$A = - \frac{3}{4} \left({x}^{2} - 160 x + 6400 - 6400\right)$

$A = - \frac{3}{4} \left({x}^{2} - 160 x + 6400\right) + 4800$

$A = - \frac{3}{4} {\left(x - 80\right)}^{2} + 4800$

The vertex is at $\left(80 , 4800\right)$.

Hence, the maximum area is given by dimensions of $80$ meters by $60$ meters, where the side cut in half is $80$ metres. The maximum area is $4800$.

Hopefully this helps!