A company manufactures closed rectangular boxes of volume 8 cubic feet. If the per square foot cost of the material for the top and bottom of a box is twice the per square foot cost of the material for the four sides of the box find the dimensions of box which minimizes the cost of material.

the dimensions of box which minimizes the cost of material are

(x, y, z) = [(∛4), (∛4), (2∛4)]

(x, y, z) = [1.587, 1.587, 3.175]

Step-by-step explanation:

Let the length, breadth and width of the box be box be x, y and z respectively.

V = xyz = 8 ft³

If the cost per unit area of the side of the rectangular box is assumed to be 1.

The cost per unit area of the top and bottom will be 2.

Area of the top and bottom = 2xy

Area of the four sides = 2xz + 2yz

Cost function for the box = (2)(2xy) + 2xz + 2yz = 4xy + 2xz + 2yz

C(x,y,z) = 4xy + 2xz + 2yz

We are to minimize this cost function subject to the constraint

Constraint = xyz - 8

We will use the Lagrange multiplier method.

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) = 4xy + 2xz + 2yz - λ(xyz - 8)

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) = 4y + 2z - λyz = 0

λ = (4y + 2z)/yz = (4/z) + (2/y)

(∂L/∂y) = 4x + 2z - λxz = 0

λ = (4x + 2z)/xz = (4/z) + (2/x)

(∂L/∂z) = 2x + 2y - λxy = 0

λ = (2x + 2y)/xy = (2/y) + (2/x)

(∂L/∂λ) = xyz - 8 = 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

(4/z) + (2/y) = (4/z) + (2/x)

(2/y) = (2/x)

y = x

Also,

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

(4/z) = (2/y)

2z = 4y

z = 2y

So, this means that, at the minimum point of the function,

y = x

z = 2y = 2x

we can then substitute this into the constraint equation or the solution of the fourth partial derivative

xyz - 8 = 0

(x)(x)(2x) - 8 = 0

2x³ - 8 = 0

x³ = 4

x = ∛4 = 1.587 ft

y = x = ∛4 = 1.587 ft

z = 2x = 2(∛4) = 3.175 ft

Hence, the dimensions of box which minimizes the cost of material are

(x, y, z) = [(∛4), (∛4), (2∛4)] = [1.587, 1.587, 3.175]

Hope this Helps!!!