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I calculated cost to find a minimum as $C=0.11xy+0.12xz+0.12yz$ for volume $xyz=668.25$. I ended up with critical points $c_x=0.11-80.19y^2=0$, and $c_y=0.11y-80.19x^2=0$ after makin the function $2$ variable by solving for z and substituting into the cost. I am now at a loss to solving these to find my critical point(s).

user141592
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Phelvrey
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  • Please, please learn to format you questions correctly. It's hard to help when the question is difficult to read. – user141592 Nov 03 '14 at 04:25
  • I apologize, I am working from a tablet with a laggy touch-screen and I have large fingers. Not much of an excuse, but it makes typing difficult. Also, thank you Sansoo for fixing this. – Phelvrey Nov 03 '14 at 04:28
  • Hmm. I made a mistake copying this down. The exponents in this problem should be negative. – Phelvrey Nov 03 '14 at 04:33

1 Answers1

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If you do not use Lagrange multipliers, consider the cost function $$C=\frac{11 x y}{100}+\frac{3 x z}{25}+\frac{3 y z}{25}$$ and eliminate $z$ from the constaint $$z=\frac{2673}{4 x y}$$ So, replacing, the cost function becomes $$C=\frac{11 x y}{100}+\frac{8019}{100 x}+\frac{8019}{100 y}$$ Taking derivatives leads to $$C'_x=\frac{11 y}{100}-\frac{8019}{100 x^2}$$ $$C'_y=\frac{11 x}{100}-\frac{8019}{100 y^2}$$ Obviously, these two derivatives cancel for $x=y=9$ and the corresponding $z$ is $z=\frac{33}{4}$.

I am sure that you can take from here.

Claude Leibovici
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