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Minimize 2x + 2y - z
Subject to x + y + z = 10

x,y,z >= 0

Okay so the answer to the problem is clearly -10 with x=0, y=0 and z=10 that's just common sense. But unfortunately I've got to show some reasoning. Can anybody show me what I'm supposed to actually get the solution?

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  • $\begingroup$ Since z is the only variable with a negative sign at the objective function it should be as large as possible. This is just $z=10$ $\endgroup$ – callculus Jan 21 '18 at 19:59
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$$2x+2y-z=2 (10-z)-z $$ $$20-3z $$

the mimimum of $20-3z $ is attained at the maximal value of $z $ which is $10$.

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  • $\begingroup$ The objective function has to be minimized. $\endgroup$ – callculus Jan 21 '18 at 19:56
  • $\begingroup$ @callculus I just edited. $\endgroup$ – hamam_Abdallah Jan 21 '18 at 19:57
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One thing to do is relax the problem. Instead of the problem posed, try solving the following problem: \begin{align} \text{minimize}\hspace{10pt} & 2x+2y-z \\ \text{subject to}\hspace{10pt} & x + y + z = 10. \end{align} This can be solved using Lagrange multipliers. Briefly, the point that minimizes $f(x,y,z)=0$ subject to $g(x,y,z)=0$ is the point where $\nabla f(x,y,z) = \lambda \nabla g(x,y,z)$. That is, the point that optimizes the problem is the one where the gradient of $f$ is a scalar multiple of the gradient of $g$.

Once you solve that problem, you'll see that the value that minimizes it also satisfies $x\geq 0$, $y\geq 0$, and $z\geq 0$. If a point solves a relaxed problem then it also minimizes the original problem. But, since it satisfies the additional constraint, it must minimize the original problem. So you're done!

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