Applied Calculus

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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 -

area=LW

L=800/W
Cost=3*2W+5L

cost=6W+4000/W

dcost/dw= 0=6-4000/W^2

W=sqrt(4000/6)

L= 800/W

• Applied Calculus -

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)=6-4000/x^2

Find the critical value
6-4000/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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