# Algebra

A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. 440 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.

-What is the shorter side of the playground? (220/3 sq ft)
-What is the longer side of the playground?
-What is the maximum area?

I know the shorter side is 220/3 sq ft, but I'm having difficulty getting the next two answers.

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1. I will assume you made a diagram.
On my diagram, I have each of the long sides as y, and each of the 3 shorter sides as x
so we have 3x + 2y = 440
y = (440 - 3x)/2 = 220 - (3/2)x

area = xy
= x(220 - (3/2)x)
= 220x - (3/2)x^2

from the subject title, I assume you don't take Calculus, then this would be easy from here on, so ....

area = 220x - (3/2)x^2 is a downwards opening parabola, thus the area will have a maximum

we need the vertex.
Easiest way to get the vertex,
the x of the vertex is -b/(2a)
= -220/-3
= 220/3 , which you had but it should be ft, not square feet

so if x = 220/3
y = 220 - (3/2)(220/3)
= 220 - 110
= 110

short side is 220/3 ft, longer side is 110 ft
area = xy
= (220/3)(110)
= 24200/3 or 8066 2/3 square feet
or appr 8066.7 ft^2

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2. No, not in calculus yet - I will take that next.

Thank you so much for your help!

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