SOLUTION: A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. 452 feet of fencing is used. Find the dimensions of the pla
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Question 1120594: A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. 452 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. I need to find the shorter side(s), longer side(l),and maximum area(m). For this I had s=113ft, l=226ft, m=25538square ft. WHat did I do wrong? Found 2 solutions by solver91311, ankor@dixie-net.com:Answer by solver91311(24713) (Show Source):
Using your variables, as the long side and as the short side, the total amount of fence is , from which we get
The area is the long side times the short side, so area as a function of the length of the short side is:
The graph of is a parabola opening downward, hence has a maximum at the vertex. The value of the independent variable at the vertex of a parabola of the form is given by:
In your case:
From that we can calculate the measure of the long side:
Then the area is just the short side times the long side:
You can do your own arithmetic.
John
My calculator said it, I believe it, that settles it
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.
452 feet of fencing is used.
Find the dimensions of the playground that maximize the total enclosed area.
:
the dividing fence will be the width also; therefore
2L + 3W = 452
2L = -3W + 452
divide by 2
L = -1.5W + 226
:
Area = L * W
Replace L with (-1.5W+ 226)
A = W(-1.5W+226)
A = -1.5W^2 + 226W
A quadratic equation, max area occurs on the axis of symmetry, x= -b/(2a)
W =
W = 75 ft is the Width
Find the Length
L = -1.5(75) + 226
L = 113 ft is the length
Find Area
A = 113 * 75
A = 8,512 sq ft