Ignoring the walls’ thickness, determine the outside dimensions that will minimize a closed box’s cost if it has a square base and top, if its volume is 32 cubic meters, and if the cost per square meter for the top, bottom, and sides respectively are $.03, $.04, and $.05.
(solve step by step please!)
3 answers
4/2
can you explain how you got that please?
let the base be x m by x m, let the height be y m
x^2 y = 32
y = 32/x^2
cost = cost of base + cost of top + cost of 4 sides
= .03x^2 + .04x^2 +4(.05)xy
= .07x^2 + .2x(32/x^2)
= .07x^2 + 6.4/x
d(cost)/dx = .14x - 6.4/x^2
= 0 for a min of cost
6.4/x^2 = .14x
x^3 = .896
x = .896^(1/3) = appr .964 m
then y = 32/.964^2 = appr 34.43 m
The answer seems unreasonable, but I just can't find any errors in my solution.
Even wrote it out on paper.
x^2 y = 32
y = 32/x^2
cost = cost of base + cost of top + cost of 4 sides
= .03x^2 + .04x^2 +4(.05)xy
= .07x^2 + .2x(32/x^2)
= .07x^2 + 6.4/x
d(cost)/dx = .14x - 6.4/x^2
= 0 for a min of cost
6.4/x^2 = .14x
x^3 = .896
x = .896^(1/3) = appr .964 m
then y = 32/.964^2 = appr 34.43 m
The answer seems unreasonable, but I just can't find any errors in my solution.
Even wrote it out on paper.
0 answers