# How do you maximize 3x+4y-yz, subject to x+y<4, z>3?

May 22, 2016

$\infty$

#### Explanation:

$f \left(x , y , z\right) = 3 x + 4 y - y z$ has not stationary points because there are no points obeying the condition

$\nabla f \left(x , y , z\right) = \vec{0}$

So their extrema could be located at the viable region frontiers. Taking a restriction frontier, for instance ${g}_{2} \left(x , y , z\right) = z - 3 = 0$ and substituting in $f \left(x , y , z\right)$ we get

$f {\left(x , y , z\right)}_{{g}_{2}} = {f}_{{g}_{2}} \left(x , y\right) = 3 x + y$

calculating $\nabla {f}_{{g}_{2}} \left(x , y\right) = \left\{3 , 1\right\}$
so also no stationary points over $z = 3$

Maximize ${f}_{{g}_{2}} \left(x , y\right) = 3 x + y$ with the border restriction
$x + y = 4$. Applying the same idea as before, substituting the border relation in the objective function, we attain
${\left({f}_{{g}_{2}}\right)}_{{g}_{1}} = 3 x + \left(4 - x\right) = 2 x + 4$ we see that the value range for ${\left({f}_{{g}_{2}}\right)}_{{g}_{1}}$is unlimited