Asked by John
A builder intends to construct a storage shed having a volume of 1000 cubic feet, a flat roof and a rectangular base whose width is three fourths the length. The cost per square foot of the material is $3 for the floor, $5 for the sides and $4 for the roof. What dimensions will minimize the cost?
Answers
Answered by
oobleck
If the base width is w, the length is 4/3 w
So, the volume, with height h is
v = w(4/3 w)h
h = 1000/(4/3 w^2) = 750/w^2
Now, the cost is
c(w) = 3(4/3 w^2) + 5(2(wh + 4/3 wh)) + 4(4/3 w^2)
= 28/3 w^2 + 17500/w
Now, minimum c will occur when dc/dw = 0
I get w=5∛(15/2)
So, the volume, with height h is
v = w(4/3 w)h
h = 1000/(4/3 w^2) = 750/w^2
Now, the cost is
c(w) = 3(4/3 w^2) + 5(2(wh + 4/3 wh)) + 4(4/3 w^2)
= 28/3 w^2 + 17500/w
Now, minimum c will occur when dc/dw = 0
I get w=5∛(15/2)
Answered by
Damon
width = w
length = 4 w/3
height = h
floor area = 4 w^2/3 so floor cost = $4 w^2
side area = h (2 w + 8w/3) = (h/3)(14 w) so side cost =$ (70/3)hw
roof cost = 4(4w^2/3) = $16 w^2 /3
total cost = (1/3) (28 w^2 + 70hw)
but
volume = 1000 = h (4 w^2/3)
so h = 750/w^2
so total cost
c = (1/3) (28 w^2 + 70hw) = (1/3) (28 w^2 + 70*750/w)
3 c = 28 w^2 + 52500/w
minimizw 3 c
d3c/dw = 0 = 56 w -52500/w^2
56 w^3 = 52500
w^3 = 937.5
w = 9.79 etc, check my arithmetic !
length = 4 w/3
height = h
floor area = 4 w^2/3 so floor cost = $4 w^2
side area = h (2 w + 8w/3) = (h/3)(14 w) so side cost =$ (70/3)hw
roof cost = 4(4w^2/3) = $16 w^2 /3
total cost = (1/3) (28 w^2 + 70hw)
but
volume = 1000 = h (4 w^2/3)
so h = 750/w^2
so total cost
c = (1/3) (28 w^2 + 70hw) = (1/3) (28 w^2 + 70*750/w)
3 c = 28 w^2 + 52500/w
minimizw 3 c
d3c/dw = 0 = 56 w -52500/w^2
56 w^3 = 52500
w^3 = 937.5
w = 9.79 etc, check my arithmetic !
Answered by
Leah
I thought it was 3/4L or is that the same as 4 w/3?
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