I have the following non linear programming to solve:
$$\left\{\begin{matrix} \min & (x-y)^2 +e^z+e^{-z} \\ \text{s.t.} & xz=0 \\ & yz=0 \end{matrix}\right.$$
The book suggests to make one or more change of variables. I couldn't find a change of variables that makes the non linear programming easier to solve.
I would be glad if someone could help me.
On
Just as CTNT suggested, let's make a new variable $w=(x-y)$. Note that while $xz=yz=0$ implies $wz=0$, the converse is not true. If we come across a 'solution' where $w=0$ and $z\neq 0$, then we have to check if $x=y=0$.
That being said, let's return to the problem. We can minimize $w^2+e^z+e^{-z}$ by setting the gradient equal to $0$:
$$\left(2w,e^z-e^{-z}\right)=(0,0)$$
So $w=0$ and $z=0$. Note that our original constraints are still satisfied. As noted by User818, this means the original function is minimized at $z=0$ along the line $x=y$. Plugging in $w=0$ and $z=0$ into the expression $w^2+e^z+e^{-z}$, we see that the minimum value is $2$.
I'm not into non-linear programming so I'm not sure what the hint is getting at. That being said, I have a solution:
We can re-write the expression to be minimized as $$f(x,y,z) = (x-y)^2 + 2 \cosh z.$$ Now $\cosh z \ge 1$ for all $z$ and $(x-y)^2 \ge 0$ for all $x,y$ so certainly $f(x,y,z) \ge 2$ for all $x,y,z$. If we can find $x,y,z$ satisfying the constraint and $f(x,y,z) = 2$ then we are done. This is quite simple just by minimizing each of $\cosh z$ and $(x-y)^2$ individually. Taking $z = 0$ and $x=y$, we see that the constraints are met and $f(x,y,z) = 2$. Thus the expression is minimized at $z = 0$ along the ling $x=y$.