SOLUTION: Nicky has 120 feet of fence to put around a rectangular garden. if a 10 foot opening is left on one side for a gate. what would be the length and width for maximum area? L=20 ft w=

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Question 1001420: Nicky has 120 feet of fence to put around a rectangular garden. if a 10 foot opening is left on one side for a gate. what would be the length and width for maximum area? L=20 ft w=20 ft
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
A square shape will give maximum area. I do not give that analysis here. The perimeter of this would be 120%2B10, to include both the fence and the gate(unfenced). Cut the 130 feet into four equal parts, and this is the side of the square garden.

32%261%2F2 feet square.


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MAXIMIZE AREA

Dimensions are x and y.
The entire perimeter of the garden is length_of_fencing PLUS 10 feet for gate; so this means the perimeter of the garden is 120%2B10=130 feet.

Two basic equations are needed.
system%282x%2B2y=130%2CA=xy%2CA=AreaOfGarden%29.

Use these to make a function A dependent on just ONE of the variables, either x or y.

2x%2B2y=130
x%2By=65
y=65-x
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A=xy
highlight%28A=x%2865-x%29%29------Area A is a function of the dimension, x, a quadratic function, and you can look for the MAXIMUM. You know there is a maximum because the coefficient on x%5E2 will be negative...
The max area will occur in the exact middle of the roots or zeros or x-intercepts of A%28x%29.
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Solve for the x-intercepts.
A%28x%29=0=x%2865-x%29

The roots are 0 and 65.
The value for x exactly in the middle is %280%2B65%29%2F2=65%2F2=32%261%2F2.

WHAT IS THE VALUE FOR y FOR THIS VALUE OF x ?
WHAT IS THE AREA AT THIS VALUE OF x ?