Question #106530
Show the COMPLETE SOLUTION for the given problem.

1. Use Lagrange Multiplier to determine the dimensions of a rectangular box, open at the top, having a volume of 32 cubic feet and requiring the least amount of material for its construction.
1
Expert's answer
2020-03-26T12:29:43-0400

We label the dimensions as l,b,h.Our objective function is of the form min  L0=lb+2bh+2lhOur constraint is on the volume lbh=32Thus, our unconstrained objective function is L=(lb+2bh+2lh)λ(32lbh)Ll=(b+2h)+λLb=(l+2h)+λLh=2(b+l)+λLλ=32lbhSetting all the partial derivatives to 0, we get 2(l+b)=(b+2h)=(l+2h)    l=b=23hSince lbh=32,  l=b=431/3,  h=232/3These are the optimum dimensions\text{We label the dimensions as }l,b,h.\\ \text{Our objective function is of the form }\\\mathrm{min}\; L_0=lb+2bh+2lh \\ \text{Our constraint is on the volume }lbh=32\\ \text{Thus, our unconstrained objective function is }\\ L=(lb+2bh+2lh)-\lambda(32-lbh)\\ \frac{\partial L}{\partial l}=(b+2h)+\lambda\\ \frac{\partial L}{\partial b}=(l+2h)+\lambda\\ \frac{\partial L}{\partial h}=2(b+l)+\lambda\\ \frac{\partial L}{\partial \lambda}=32-lbh\\ \text{Setting all the partial derivatives to 0, we get }\\ 2(l+b)=(b+2h)=(l+2h)\implies l=b=\frac{2}{3}h\\ \text{Since }lbh=32, \; l=b=\frac{4}{3^{1/3}}, \; h=2*3^{2/3}\\ \text{These are the optimum dimensions}


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