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I am working on a general LP exercise, and am having issues with proving the following:

I have a general minimization problem that looks like this:

$$\text{Minimum Problem} = \begin{cases} \text{min } \ \ \ C(r,h) \\ \text{s.t.:}\ \ SA(r,h)>0\\ \ \ \ \ \ \ \ \ \ \ \ C(r,h)>0\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r>0\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h>0\\ \ \ \ \ \ \ \ \ \ \ \ V(r,h)=20 \end{cases} $$

and want to show it as a general maximization problem, to which I believe looks like:

$$\text{Maximum Problem} = \begin{cases} \text{max } \ -C(r,h) \\ \text{s.t.:}\ \ \ \ \ SA(r,h)>0\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C(r,h)>0\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r>0\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h>0\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V(r,h)= 20 \end{cases} $$

But (if this is right) I don't know how to prove it or how to show it can't be done... any ideas would be so helpful.

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    You don't need to negate your constraints or switch their signs. Simply take your minimization problem, negate it then solve. Then take the negation of the answer, i.e. in general $z=\min f(x) s.t. x \in \mathcal{X} = - \max -f(x) s.t. x \in \mathcal{X}$. – 2017-02-02
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    @David, fixed, I think? – 2017-02-03
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    As formulated, those problems will have the same optimal decision variables. If you want them to have the same optimal value, you need to multiply the final solution to one of the problems by -1. That is, the -1 needs to be outside the max or min function itself. – 2017-02-03

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