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.

Expert's answer

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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