Asked by J.
An open water tank with a square base is to be made from a thin sheet of metal.
find the length of the square base and the height of the tank so that the least amount of metal is used to make a tank of capacity 8 meters cubed
find the length of the square base and the height of the tank so that the least amount of metal is used to make a tank of capacity 8 meters cubed
Answers
Answered by
Reiny
let the base be x m by x m , and let the height be y m
V= x^2y
8 = x^2y ---> y = 8/x^2
Surface area
= SA = x^2 + 4xy
= x^2 + 4x(8/x^2)
= x^2 + 32/x
d(SA)/dx = 2x - 32/x^2
= 0 for a min of SA
2x = 32/x^2
x^3 = 16
x = 16^(1/3) = appr 2.52 m
y = .....
V= x^2y
8 = x^2y ---> y = 8/x^2
Surface area
= SA = x^2 + 4xy
= x^2 + 4x(8/x^2)
= x^2 + 32/x
d(SA)/dx = 2x - 32/x^2
= 0 for a min of SA
2x = 32/x^2
x^3 = 16
x = 16^(1/3) = appr 2.52 m
y = .....
Answered by
Josh
Your solution for the SA is wrong which makes the whole equation wrong. The formula for surface area is
2x² + 4xy
Just follow the steps you did then you'll arrive at the answer of
x = 2
h = 2
2x² + 4xy
Just follow the steps you did then you'll arrive at the answer of
x = 2
h = 2
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