Question 58256: A farmer has 96 meters of fencing. He wants to enclose a rectangular field and build a fence across the middle, parallel to the shorter ends.
a. Express the total area of the field as a function of the width w, where w is the length of the shorter ends (hint: result is a quadratic)
b. Algebraically determine the value of x for which the area is maximum.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A farmer has 96 meters of fencing. He wants to enclose a rectangular field and build a fence across the middle, parallel to the shorter ends.
:
a. Express the total area of the field as a function of the width w, where w is the length of the shorter ends (hint: result is a quadratic)
:
Because of that fence down the middle, the perimeter is:
2L + 3W = 96
2L = 96 - 3W
L = (48 - 1.5W); divided eq by 2.
:
Area = L*W
Let y = area
Substitute (48-1.5W) for L
y = (48 - 1.5W) * W
y = 48W - 1.5W^2
Arrange as a Quadraic eq: y = -1.5W^2 + 48W
b. Algebraically determine the value of x for which the area is maximum.
:
Use the vertex equation: x = -b/(2a); in our eq, a=-1.5, b= 48
:
x = -48/(2*-1.5)
x = -48/-3
x = +16 meters is the width for max area
:
Find the max area, substitute 16 for x in the equation: y = -1.5x^2 + 48x
:
y = -1.5(16^2) + 48(16)
y = -1.5(256) + 768
y = -384 + 768
y = +384 sq meters is max area and occurs when the width is 16 meters.
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