SOLUTION: Any help I could get on this problem would be greatly appreciated! I cannot figure it out! A rectangular playground is to be fenced off and divided in two by another fence para

Question 720082: Any help I could get on this problem would be greatly appreciated! I cannot figure it out!
A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. 472 feet of fencing is used. Find the dimensions of the playground that maximize the total enclosed area. Remember to reduce any fractions and simplify your answers as much as possible.

Thank you so much for any help!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground.
472 feet of fencing is used.
Find the dimensions of the playground that maximize the total enclosed area.
:
Normally the perimeter would be 2L + 2W but since we have one more side to divide the area we will write it:
2L + 3W = 472
divide by 2
L + 1.5W = 236
then
L = (236-1.5w)
:
The area: A = L * W
replace L with (236-1.5W)
A = W(236-1.5W)
A = -1.5W^2 + 236W
This is a quadratic equation, if we find the axis of symmetry, we will know the value of W that gives max area.
Formula for the axis of symmetry: x = -b/2a, in this problem it would be
W = -236/(2*-1.5)
W = -236/-3
W = +78 ft is the width
Find the Length
L = 236 - 1.5(78.667)
L = 159 is the length for max area
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