Answer:
x = y = ∛(2V/5)
z = ∛(25V/4)
Explanation:
We want to minimize the cost as a function of x, y, and z subject to the constraint that xyz = V. Using Lagrange multipliers, we can write the Lagrangian as ...
L = 2z(x+y) +5xy + λ(xyz -V)
The partial derivatives are all zero at the extreme:
∂L/∂x = 0 = 2z +5y +λyz
∂L/∂y = 0 = 2z +5x +λxz
∂L/∂z = 0 = 2(x+y) +λxy
∂L/∂λ = 0 = xyz -V
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Solving each of the first three equations for λ, we get ...
λ = -(2/y +5/z)
λ = -(2/x +5/z)
λ = -2(1/y +1/x)
Subtracting the second from the first, we get ...
0 = -(2/y +5/z) +(2/x +5/z)
2/y = 2/x
y = x
In terms of x, then the third equation tells us ...
λ = -4/x
Using that in the second equation gives ...
-4/x = -2/x -5/z
2/x = 5/z
z = (5/2)x
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Now, in terms of x, we can rewrite the original ∂L/∂λ as ...
(x)(x)(5/2x) - V = 0
x³ = (2/5)V
x = y = ∛(2V/5)
z = (5/2)x
z = ∛(25V/4)
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The minimum cost aquarium has a square base and is of such dimensions that each pair of opposite sides has the same cost as the base.