Answer to Question #327549 in Calculus for math

Question

A rectangular field is to be enclosed by a fence then subdivided into three areas by fences

parallel to the shorter side. Find the dimensions of the field with the largest total area that can be

enclosed with 260 m of fencing. What is the area?

Answer 1

As rectangular field is divided into three areas the fence is consist of 4 short sides and 2 large sides.

Define $x$ m is a short side. Then $(260-4x)/2$ m is a large side. Total area:

$S(x)=x(260-4x)/2$

Let’s find x for which the total area is maximum:

$S’(x)=130-4x=0$

$x=32.5$

As $S’(30)=130-120>0$ and $S’(35)=130-140<0$ we can conclude $x=32.5$ corresponds to maximum value $S(x)$.

Sides of the rectangular field:

32.5 m and $(260-4\cdot32.5)/2=65$ m.

Largest total area:

$S=32.5\cdot65=2112.5$ $m^2$