Design a rectangular milk carton box of width $$w, length $$l, and height $$h, which holds $$128 cubic cm of milk. The sides of the box cost $$1 cent per square cm and the top and bottom cost $$2 cents per square cm. Find the dimensions of the box that minimize the total cost of materials used.

Length = 4cm

Width = 4cm

Height = 8cm

Step-by-step explanation:

The volume of the box = 128cm^3

LWH = Volume

LWH = 128cm^3

The side of the box = $1 per cm^2

The top and bottom of the box = $2 per cm^2

Let C be the cost function

C (LWH) = (1) 2H (L+W) + (2) 2LW

from LWH = 128cm^39

H = 128/LW

put H = 128/LW in equation for C (LWH)

C (LW) = (1) 2 (128/LW) + (L+W) + (2) 2LW

= 256/LW (L+W) + 4LW

= 256 (1/L + 1/W) + 4LW

Differentiate C with respect to L

dC/dL = 4W - 256/L^2 = 0

Differentiate C with respect to W

dC/dW = 4L - 256/W^2 = 0

The cost is minimum when the two partial derivatives equal 0

From 4W - 256/L^2 = 0

4W = 256/L^2

W = (256/L^2) 1/4

W = 64/L^2

From 4L - 256/W^2 = 0

4L = 256/W^2

L = (256/W^2) 1/4

L = 64/W^2

Since L = W,

L = W = cuberoot (64)

L = W = 4cm

Recall that H = 128/LW

H = 128 / (4*4)

H = 128/16

H = 8cm

therefore;

L = 4cm

B = 4cm

H = 8cm