SOLUTION: Need help Please!!!! A rancher with 750ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. a

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Question 181262: Need help Please!!!!
A rancher with 750ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle.
a. Find a function that models the total area of the four pens.
b. Find the largest possible total area of the four pens.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A rancher with 750ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle.
:
If you draw this, you can see that the fencing required: 2L + 5W = 750
:
a. Find a function that models the total area of the four pens.
:
Let x = the width
then
2L + 5x = 750
2L = (750 - 5x)
L = %28750-5x%29%2F2
L = (375 - 2.5x)
:
Area = x * L
Substitute (375-2.5x) for L
A = x*(375-2.5x)
A = 375x - 2.5x^2
the function that models the total area
f(x) = -2.5x^2 + 375x
:
:
b. Find the largest possible total area of the four pens.
:
Since this is a quadratic equation, we can do this by finding the axis
of symmetry, then find the vertex. Formula for that is: x = %28-b%29%2F%282a%29
x = %28-375%29%2F%282%2A-2.5%29
x = %28-375%29%2F%28-5%29
x = +75, the max area will be when width is 75 ft
;
Find the vertex (max area):
A = -2.5(75^2) + 375(75)
A = -2.5(5625) + 28125
A = -14062.5 + 28125
A = +14062.5 sq ft is the max area
;
Did this make sense to you, any questions?