# How do you minimize and maximize f(x,y)=x+y constrained to 0<x+3y<2?

Aug 1, 2016

This problem is unbounded.

#### Explanation:

The linear function to maximize/minimize $f \left(x , y\right)$ grows
in the direction of its gradient

${\nabla}_{f} = \nabla f \left(x , y\right) = \left\{\frac{\partial f}{\partial x} , \frac{\partial f}{\partial y}\right\} = \left\{1 , 1\right\}$

the linear restrictions offer boundaries at

${g}_{1} \left(x , y\right) = x + 3 y = 0$
${g}_{2} \left(x , y\right) = x + 3 y = 2$

with constant declivity given by the vector

${\vec{r}}_{1} = {\vec{r}}_{2} = \vec{r} = \left\{3 , - 1\right\}$. Note that the declivity vector is normal to the restriction gradient vector given by

${\nabla}_{{g}_{1}} = {\nabla}_{{g}_{1}} = {\nabla}_{g} = \left\{1 , 3\right\}$

Concluding, the projection of ${\nabla}_{f}$ onto $\vec{r}$ is

$\frac{\left\langle{\nabla}_{f} , \vec{r}\right\rangle}{\left\lVert \vec{r} \right\rVert} = \frac{\left\langle\left\{1 , 1\right\} , \left\{3 , - 1\right\}\right\rangle}{\left\lVert \left\{- 3 , 1\right\} \right\rVert} = \frac{2}{\sqrt{10}} = {C}^{t e}$ so $f \left(x , y\right)$ keeps growing or decreasing along those boundaries.