SOLUTION: Rock had 800 foot fence material. He wants to form a rectangular closure or casing and then divides it into 3 section by running 2 lengths of fence parallelized to one side. What m

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Question 1082918: Rock had 800 foot fence material. He wants to form a rectangular closure or casing and then divides it into 3 section by running 2 lengths of fence parallelized to one side. What must be the dimension of the enclosure be in order to maximize the enclosed area?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Rock had 800 foot fence material.
He wants to form a rectangular closure or casing and then divides it into 3 section by running 2 lengths of fence parallelized to one side.
What must be the dimension of the enclosure be in order to maximize the enclosed area?
:
let L = the length of the enclosure
let w = the width and the 2 dividing fences
;
2L + 4w = 800
simplify, divide by 2
L + 2w = 400
L = 400-2w, use this form for substitution
:
Area = L * w
replace L with (400-2w)
A = (400-x)*2w
A quadratic equation
A = -2w^2 + 400w
Max area occurs on the axis of symmetry, find that using x =-b/(2a)
w = %28-400%29%2F%282%2A-2%29
w = +100 ft is the width for max area
and
L = 400-2(100)
L = 200 ft is the length