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?
On
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!
$$2x+2y-z=2 (10-z)-z $$ $$20-3z $$
the mimimum of $20-3z $ is attained at the maximal value of $z $ which is $10$.