Math lagrange multiple

Question

The material for a rectangular box costs RM 3 per square foot for the bottom and

RM 2 per square foot for the sides and top. Find the dimensions of the container so

that its volume is 12 cubic feet and the cost is minimum. Use Lagrange Multipliers to

solve this problem and give your answer in two decimal places

Answer 1

Bottom and top have area lw, 2 sides have area lh, and 2 sides have area wh.

Thus, as the bottom cost is 3 and all others are 2, cost is 3lw + 2lw + 2(2lh) + 2(2wh)

max 5lw + 4lh +4wh subject to lwh = 12

Using LaGrange multiplier theory, we have

5lw + 4lh +4wh + lambda ( lwh - 12)

Taking derivatives and setting equal to 0, we have

5w + 4h + lambda wh = 0 Thus, 5w + 4h = -lambda wh; -lambda = (5w+4h)/wh

5l + 4h + lambda lh = 0 ; - lambda = (5l + 4h)/lh

4l + 4w + lambda lw = 0 -lambda = (4l+4w)/lw

Setting - lambdas equal,

Then, (5w + 4h)/wh = (5l + 4h)/lh = (4l + 4w)/lw

5/h + 4/w = 5/h + 4/l = 4/w + 4/l

Thus, 4/w = 4/l and 5/h = 4/w, or w = l and h = 5/4 w

Thus, as the volume = lwh = 12, lw (5/4 w) = 12, or 5/4 w^3 = 12

w^3 = 9.6

w = l = 9.6^(1/3) = 2.13 h = 5/4 w = 2.66

Cost is 5 * (9.6)^(2/3) + 8 * 5/4 * (9.6)^(2/3) = 15 * (9.6)^(2/3) = 67.75

(Note that, if the three dimensions were all 12^(1/3), the cost would be 13 * 12^(2/3) = 68.14)