Question

a rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Eight hundred and forty feet of fencing is used. Find the dimension of the playground that maximize the total enclosed area. What is the maximum area?

Answer 1

We will do first the diagram

total fence used 840 ft

The total fence in terms of the variables

3y+2x=840

then we isolate the y

$y=\frac{840-2x}{3}$

then we calculate the area

$A=x\cdot\frac{840-2x}{3}=\frac{840x-2x^2}{3}$

then we have a quadratic equation, which means that the max will be the vertex

the

x coordinate of the vertex can be calculated

$x=\frac{-b}{2a}$

where

a=-2

b=840

we substitute the values

$x=\frac{-840}{2(-2)}=\frac{-840}{-4}=210$

then we susbtitute in the function of the area

$\begin{gathered} A=\frac{840(210)-2(210)^2}{3} \\ A=29400 \end{gathered}$

then we find y

$A=x\cdot y$$y=\frac{A}{x}=140$

The dimensions that maximize the enclosed area

x=210 ft

y=140ft

The maximum area is 29400 ft^2