Answer to Question #304525 in Calculus for Anniee

Question

A farmer has 800m of fencing material to be enclose a rectangular pen adjacent to a long-existing wall. He will use the wall for one side of the pen and the available fencing material for the remaining three sides. What is the maximum area that can be enclosed this way?

Answer 1

Let $x$ denote the width of a rectangular pen, $y$ denote the length of a rectangular pen.

Given

$$2x+y=800$$

Then

$$y=800-2x$$

The area that can be enclosed is $A=xy.$ Substitute

$$A(x)=x(800-2x), 0<x<400$$

Find the first derivative with respect to $x$

$$A'(x)=(x(800-2x))'=800-4x$$

Find the critical number(s)

$$A'(x)=0=>800-4x=0$$

$$x=200$$

Critical number: $x=200.$

If $0<x<200, A'(x)>0, A(x)$ increases.

If $200<x<400, A'(x)<0, A(x)$ decreases.

The function $A(x)$ has the absolute maximum on $[0, 400]$ at $x=200.$

$$y=800-2(200)=400$$

$$A=200(400)=80000(m^2)$$

The maximum area that can be enclosed this way is $80000 \ m^2.$