The base of an aquarium with given volume V is made of slate and the sides are made of glass. If the slate costs seven times as much (per unit area) as glass, use Lagrange multipliers to find the dimensions of the aquarium that minimize the cost of the materials. (Enter the dimensions as a comma separated list.)

x = ∛(2V/7)

y = ∛(2V/7)

z = 3.5 [∛(2V/7)]

{x,y,z} = { ∛(2V/7), ∛(2V/7), 3.5[∛(2V/7)] }

Step-by-step explanation:

The aquarium is a cuboid open at the top.

Let the dimensions of the base of the aquarium be x and y.

The height of the aquarium is then z.

The volume of the aquarium is then

V = xyz

Area of the base of the aquarium = xy

Area of the other faces = 2xz + 2yz

The problem is to now minimize the value of the cost function.

The cost of the area of the base per area is seven times the cost of any other face per area.

With the right assumption that the cost of the other faces per area is 1 currency units, then, the cost of the base of the aquarium per area would then be 7 currency units.

Cost of the base of the aquarium = 7xy

cost of the other faces = 2xz + 2yz

Total cost function = 7xy + 2xz + 2yz

C(x,y,z) = 7xy + 2xz + 2yz

We're to minimize this function subject to the constraint that

xyz = V

The constraint can be rewritten as

xyz - V = 0

Using Lagrange multiplier, we then write the equation in Lagrange form

Lagrange function = Function - λ(constraint)

where λ = Lagrange factor, which can be a function of x, y and z

L(x,y,z) = 7xy + 2xz + 2yz - λ(xyz - V)

We then take the partial derivatives of the Lagrange function with respect to x, y, z and λ. Because these are turning points and at the turning point, each of the partial derivatives is equal to 0.

(∂L/∂x) = 7y + 2z - λyz = 0

λ = (7y + 2z)/yz = (7/z) + (2/y) (eqn 1)

(∂L/∂y) = 7x + 2z - λxz = 0

λ = (7x + 2z)/xz = (7/z) + (2/x) (eqn 2)

(∂L/∂z) = 2x + 2y - λxy = 0

λ = (2x + 2y)/xy = (2/y) + (2/x) (eqn 3)

(∂L/∂λ) = xyz - V = 0

We can then equate the values of λ from the first 3 partial derivatives and solve for the values of x, y and z

(eqn 1) = (eqn 2)

(7/z) + (2/y) = (7/z) + (2/x)

(2/y) = (2/x)

y = x

Also,

(eqn 1) = (eqn 3)

(7/z) + (2/x) = (2/y) + (2/x)

(7/z) = (2/y)

z = (7y/2)

Hence, at the point where the box has minimal area,

y = x,

z = (7y/2) = (7x/2)

We can then substitute those into the constraint equation for y and z

xyz = V

x(x)(7x/2) = V

(7x³/2) = V

x³ = (2V/7)

x = ∛(2V/7)

y = x = ∛(2V/7)

z = (7x/2) = 3.5 [∛(2V/7)]

The values of x, y and z in terms of the volume that minimizes the cost function are

{x,y,z} = {∛(2V/7), ∛(2V/7), 3.5[∛(2V/7)]}

Hope this Helps!!!