Posted by Trisha on Sunday, April 8, 2012 at 7:02pm.
A hardware store owner chooses to enclose an 800 square foot rectangular area in front of her store so that one of the sides of the store will be used as one of the four sides of the fence. If the two sides that come out from the store front cost $3 per running foot for materials and the side parallel to the store front costs $5 per running foot for materials, then find the dimensions of the fence that will minimize the cost to construct the fence. Round all dimensions to the nearest foot.

Applied Calculus  bobpursley, Sunday, April 8, 2012 at 7:25pm
area=LW
L=800/W
Cost=3*2W+5L
cost=6W+4000/W
dcost/dw= 0=64000/W^2
W=sqrt(4000/6)
L= 800/W

Applied Calculus  Trisha, Sunday, April 8, 2012 at 7:56pm
Thank you, I got a very similar trying to do it the way our homework assignments were done. But your way seems much simpler. I got 25.8198898 by 30.76923077.
C=3x+3x+5y
=6x+5y
xy=800
y=800/x
c(x)=6x+5(800/x)
=6x+4000/x
c'(x)=64000/x^2
Find the critical value
64000/x^2=0
6=4000/x^2
6x^2=4000
x^2=4000/6
x^2=666.6666667
x=25.81988898
c"(x)=8000/x^3
C"(26)=8000/26^3
=.4551661356
xy=800
26y=800
y=800/26
y=30.76923077
The dimensions would by 26ft by 31ft.
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