The base of an aquarium with given volume v is made of slate and the sides are made of glass. if slate costs five times as much (per unit area) as glass, find the dimensions of the aquarium that minimize the cost of the materials. (assume x is the length and y is width of the base; z is the height of the aquarium. enter your answer in terms of v.

  x = y = ∛(2V/5)

  z = ∛(25V/4)

Step-by-step 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)

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.

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step-by-step explanation:

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a variable is any characteristics, number, or quantity that can be measured or counted. a variable may also be called a data item. hope i !