a box is to be constructed with a total volume of 6400 cm^3. The sides cost $5 per cm^2, the top and bottom cost $4 per cm^2 . What are the dimensions that have the correct volume, but minimize cost?

Answer 1

Let x, y, and z be the lengths of the sides of the box, with x being the height.

The volume is xyz, and we are told that this product is 6400.

The total cost of the sides will be $5*( 2xy + 2xz ) = 10x(y + z)

The cost of the top and bottom will be $4*( 2yz ) = 8yz

Let C(x,y,z) represent the total cost of the box. Then

C(x,y,z) = 10x(y + z) + 8yz

The constraints are as follows:

x > 0

y > 0

z > 0

xyz = 6400

We seek to minimize the Cost:

C(x,y,z) = 10x(y + z) + 8yz

We proceed by employing a Lagrange Multiplier, m:

L(x,y,z) = C(x,y,z) - m(6400 - xyz)

We take the partial derivatives of L with respect to x, y, and z, setting each to zero:

L_x = 10(y+z) + 0 - m(-yz) = 0

L_y = 10x(1) + 8z - m(-xz) = 0

L_z = 10x(1) + 8y - m(xy) = 0

There are common terms that can be eliminated:

10(y+z) + m(yz) = 0

10x + 8z + m(xz) = 0

10x + 8y + m(xy) = 0

Solving for -m leads to

-m = 10/z + 10/y = 10/z + 8/x = 10/y + 8/x

From which we conclude that

10/y = 8/x

10/z = 10/y

Finally,

z = y = 4x/5

Since xyz = 6400

x (5x/4) ( 5x/4) = 6400

x³ = (6400)*(4/5)*(4/5) = (4*25*8*8)(4/5)(4/5) = 4³*8² =

x = 4*4 = 16 cm

y = 5x/4 = 20 cm

z = 20 cm

Though not requested,

C(16,20,20) = 10x(y+z) + 8yz = 10(16)(20+20) + 8(20)(20) = 160*40 + 8*400 = 6400 + 3200 = 9600 cents

or $96