Asked by Anonymous
If Steve has 120 feet of fence to make a rectangular kennel and he will use his house on one side, how would the dimensions be found for a kennel with the greatest area be found?
Answers
Answered by
Damon
L + 2 w = 120
A = L w
A = (120 -2w)w = 120 w - 2 w^2
that is parabola opening down, find the vertex (I assume you do not know calculus which would be easier)
-2 w^2 + 120 w = A
complete square
w^2 - 60 w = -A/2
w^2 - 60 w + 30^2 = -A/2 + 900
(w-30)^2 = -A/2 + 900
so
w = 30
then L = 120 - 60 = 60
A = L w
A = (120 -2w)w = 120 w - 2 w^2
that is parabola opening down, find the vertex (I assume you do not know calculus which would be easier)
-2 w^2 + 120 w = A
complete square
w^2 - 60 w = -A/2
w^2 - 60 w + 30^2 = -A/2 + 900
(w-30)^2 = -A/2 + 900
so
w = 30
then L = 120 - 60 = 60
Answered by
bobpursley
Area= LW
Perimeter= 2L+ W
dArea/dL= W + dW/dl
the last term you get from taking the derivative of the perimeter equation
0 (perimeter is constant)=2 + dW/dl
so put that in the darea/dl, set to zero, and solve for W.
Perimeter= 2L+ W
dArea/dL= W + dW/dl
the last term you get from taking the derivative of the perimeter equation
0 (perimeter is constant)=2 + dW/dl
so put that in the darea/dl, set to zero, and solve for W.
Answered by
Anonymous
Thanks
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