Question 1209187: A homeowner has 80 feet of fence to enclose a rectangular garden. What dimensions for the garden give the maximum area?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: 20 feet by 20 feet
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Short explanation:
Given some amount of fencing P, the dimensions of the max area rectangle (which turns out to be a square) is P/4 by P/4
We have P = 80 feet of fencing lead to P/4 = 80/4 = 20 which is the dimensions of the square.
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Longer explanation
x = length
perimeter of a rectangle = 2*width+2*length
80 = 2*width+2*length
40 = width + length ........ divide both sides by 2
width = 40-length
width = 40-x
In short,
length = x
width = 40-x
which leads to:
area = length*width
area = x*(40-x)
area = 40x-x^2
area = -x^2+40x
Compare the equation y = -x^2+40x with the template y = ax^2+bx+c
a = -1, b = 40, c = 0
The vertex (h,k) is the highest point of this parabola.
This is because a = -1 is negative. The parabola opens downward.
Therefore, finding the vertex will help us max out the area.
h = -b/(2a)
h = -40/(2*(-1))
h = 20
This is the x coordinate of the vertex.
You can use a graphing tool like Desmos and GeoGebra to verify.
The area maxes out when the length is x = 20 feet.
The width is 40-x = 40-20 = 20 feet.
The dimensions are 20 feet by 20 feet.
Extra info: The area is 20*20 = 400 square feet.
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Another approach.
Factor -x^2+40x = 0 to get -x(x-40) = 0
From here we can quickly see that the roots are x = 0 and x = 40.
It turns out that the x coordinate of the vertex is the midpoint of these roots.
This is due to the parabola's mirror symmetry.
Add the roots and divide in half: (0+40)/2 = 20
The x coordinate of the vertex is x = 20.
The length is x = 20 and the width is 40-x = 40-20 = 20.
We have a 20 by 20 square.
Side note: you could use the quadratic formula to solve -x^2+40x=0, but it would be overkill in my opinion.
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