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

Nov 7, 2016

$\pm \infty$

#### Explanation:

${\lim}_{x \to \infty} f \left(x , x - \frac{1}{2}\right) = {\lim}_{x \to \infty} \left(\frac{x}{x - \frac{1}{2}} - x \cdot \left(x - \frac{1}{2}\right)\right) = - \infty$
${\lim}_{y \to {0}^{+}} f \left(y + \frac{1}{2} , y\right) = {\lim}_{y \to {0}^{+}} \frac{y + \frac{1}{2}}{y} - \left(y + \frac{1}{2}\right) y = + \infty$

Both choices respect the constraint.

Nov 7, 2016

There is a local máxima at $x = 0 , y = - 1$

#### Explanation:

Defining

$f \left(x , y\right) = \frac{x}{y} - x y$

and

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

the local maximization/minimization problem with inequality restrictions is transformed into an equivalent one, now with equality restrictions, so we can apply the Lagrange Multipliers technique for its resolution. The lagrangian is

$L \left(X , S , \Lambda\right) = f \left(x , y\right) + {\lambda}_{1} {g}_{1} \left(x , y , {s}_{1}\right) + {\lambda}_{2} {g}_{2} \left(x , y , {s}_{2}\right)$

Here $X = \left(x , y\right) , S = \left({s}_{1} , {s}_{2}\right) , \Lambda = \left({\lambda}_{1} , {\lambda}_{2}\right)$

The stationary points are solutions of

$\nabla L = \vec{0}$

or

{ (1/y - y + lambda_1 + lambda_2=0), ( x + x/y^2+ lambda_1 + lambda_2 = 0), (2 lambda_1 s_1 = 0), (2 lambda_2 s_2 = 0), ( x - y -s_1^2 = 0), ( x - y -1 + s_2^2= 0) :}

Solving for $X , S , \Lambda$ we obtain

$\left(x = 0 , y = - 1 , {s}_{1} = \pm 1 , {s}_{2} = 0 , {\lambda}_{1} = 0 , {\lambda}_{2} = 0\right)$

Here we observe that ${s}_{2} = 0$ so the active restriction is

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

Also we have

$\left(f \circ {g}_{2}\right) \left(x\right) = 1 + \frac{1}{x - 1} + x - {x}^{2}$ and

$\frac{{d}^{2}}{{\mathrm{dx}}^{2}} \left(f \circ {g}_{2}\right) \left(x\right) = \frac{2}{x - 1} ^ 3 - 2$ and
$\frac{{d}^{2}}{{\mathrm{dx}}^{2}} \left(f \circ {g}_{2}\right) \left(0\right) = - 4$

This result shows that the point $x = 0 , y = - 1$ is a local máxima for the problem.