# How do you maximize and minimize f(x,y)=x^2-y/x constrained to 0<=x+y<=1?

Feb 16, 2016

There is no upper bound for $f$.

The lower bound is $f \left(0 , 0\right) = 1$

#### Explanation:

Substitute $x + y = u$.

Now, optimize

$g \left(x , u\right) = {x}^{2} - \frac{u - x}{x}$

$= {x}^{2} + 1 - \frac{u}{x}$

With the only condition being $0 \le u \le 1$.

When $x$ is large, the ${x}^{2}$ term dominates and the $\frac{u}{x}$ term becomes insignificant. By increasing $x$, $f$ can be made arbitrarily large.

$\frac{\partial g}{\partial x} = 2 x + \frac{u}{x} ^ 2$

From the first equation, we see that for a given value of $u$, minimum occurs when $\frac{\partial g}{\partial x} = 0$. Therefore,

$x = - \sqrt[3]{\frac{u}{2}}$

So to find the minimum of $g$, we replace all instances of $x$ with $- \sqrt[3]{\frac{u}{2}}$.

$g \left(u\right) = {\left(- \sqrt[3]{\frac{u}{2}}\right)}^{2} + 1 - \frac{u}{- \sqrt[3]{\frac{u}{2}}}$

$= {2}^{- \frac{2}{3}} \cdot {u}^{\frac{2}{3}} + 1 + \sqrt[3]{2} \cdot {u}^{\frac{2}{3}}$

$= 3 \sqrt[3]{{u}^{2} / 4} + 1$

Even without differentiating, it is quite easy to see that $g$ is increasing for $u \in \left[0 , 1\right]$

The mimimum corresponds to $u = 0$, and consequently, $x = 0$, and $y = 0$.

The absolute minimum is $f \left(0 , 0\right) = 1$