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.

Answer:Length = 4cmWidth = 4cmHeight = 8cmStep-by-step explanation:The volume of the box = 128cm^3LWH = VolumeLWH = 128cm^3The side of the box = $1 per cm^2The top and bottom of the box = $2 per cm^2Let C be the cost function C(LWH) = (1) 2H (L+W) + (2) 2LWfrom LWH = 128cm^39H = 128/LWput 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) + 4LWDifferentiate C with respect to LdC/dL = 4W - 256/L^2 = 0Differentiate C with respect to WdC/dW = 4L - 256/W^2 = 0The cost is minimum when the two partial derivatives equal 0From 4W - 256/L^2 = 04W = 256/L^2W = (256/L^2) 1/4W = 64/L^2From 4L - 256/W^2 = 04L = 256/W^2L = (256/W^2) 1/4L = 64/W^2Since L = W, L= W = cuberoot (64)L = W = 4cmRecall that H= 128/LWH = 128/(4*4)H= 128/16H= 8cmtherefore; L= 4cmB= 4cmH= 8cm