Question
A homeowner wants to fence a rectangular garden using 60 ft of fencing. An existing stone wall will be
used as one side of the rectangle. Find the dimensions for which the area is a maximum
used as one side of the rectangle. Find the dimensions for which the area is a maximum
Answers
60 = L + 2 w so L = 60 -2 w
A = L w
A = (60-2 w)w = 60 w - 2 w^2
If you know calculus, just take derivative and set to zero and you are done
If not, complete the square to find vertex of parabola.
w^2 - 30 w = -A/2
w^2 - 30 w + 225 = -A/2 + 225
(w-15)^2 = -(1/2)(A - 450)
so
w = 15
L = 60 - 30 = 30
area = 450 by the way
A = L w
A = (60-2 w)w = 60 w - 2 w^2
If you know calculus, just take derivative and set to zero and you are done
If not, complete the square to find vertex of parabola.
w^2 - 30 w = -A/2
w^2 - 30 w + 225 = -A/2 + 225
(w-15)^2 = -(1/2)(A - 450)
so
w = 15
L = 60 - 30 = 30
area = 450 by the way
So the maximum area would be 450?
I'm trying to read through your work so I can understand how you got the answer
I'm trying to read through your work so I can understand how you got the answer
Nope I still don't get it. I understand right up until this...
"
w^2 - 30 w = -A/2
w^2 - 30 w + 225 = -A/2 + 225
(w-15)^2 = -(1/2)(A - 450)
so
w = 15
L = 60 - 30 = 30
area = 450 by the way
"
Thanks anyway!
"
w^2 - 30 w = -A/2
w^2 - 30 w + 225 = -A/2 + 225
(w-15)^2 = -(1/2)(A - 450)
so
w = 15
L = 60 - 30 = 30
area = 450 by the way
"
Thanks anyway!
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