Asked by Parveen
John is fencing a larger rectangular pen and creating 3 smaller pens of equal area (2 lengths and 4 widths). John wants to maximize the area contained by the pens. John plans on using 1200m of fencing. What is the area and dimension of the 3 smaller pens? The larger pen?
Answers
Answered by
Reiny
Length ---- y
each of the 4 width --- x
so 2y + 4x = 1200
y = 600 - 2x
area of whole pen = xy = x(600-2x) = 600x - 2x^2
d(area...)/dx = 600 - 4x
= 0 for a max of area
4x = 600
x = 150
take over
each of the 4 width --- x
so 2y + 4x = 1200
y = 600 - 2x
area of whole pen = xy = x(600-2x) = 600x - 2x^2
d(area...)/dx = 600 - 4x
= 0 for a max of area
4x = 600
x = 150
take over
Answered by
Damon
if you have not had calculus
A = 600 x - 2x^2
x^2 - 300 x = -A/2
find vertex of parabola (the top since it opens down)
complete square
x^2 - 300 x + (300/2)^2 = -A/2 + 150^2
(x-150)^2 = -(1/2)(A - 45000)
so max at x =150, area = 45000
A = 600 x - 2x^2
x^2 - 300 x = -A/2
find vertex of parabola (the top since it opens down)
complete square
x^2 - 300 x + (300/2)^2 = -A/2 + 150^2
(x-150)^2 = -(1/2)(A - 45000)
so max at x =150, area = 45000
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