\(\text{What's that? Lagrange multipliers?}\\ \text{The method of Lagrange multipliers involves solving}\\ \nabla (f(s,h) - \lambda (c(s,h)-C))\\ \text{In this problem }\\ f(s,h) = vol(s,h) = s^2 h\\ c(s,h)=6sh+6s^2 (\text{see the previous answer})\\ C=15\)
\(\text{our Lagrange equation becomes}\\ \nabla \left(s^2 h - \lambda (6sh-6s^2 - 15)\right)= (0,0,0)\\ \text{this leads to the system of equations }\\ 2sh - \lambda(6h - 12s)=0\\ s^2 - 6s\lambda = 0\\ 6sh-6s^2-15 = 0\\\)
\(\text{and these are solved for }s,~h,~\lambda\\ \text{This seems like a lot more work than the previous answer. Why do it?}\\ \text{In most problems you can't cleanly solve for one in terms of the other and substitute like we did}\)
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