# How do you maximize and minimize f(x,y)=x^2+3xy+9y^2 constrained to 0<x+3y<2?

Jun 5, 2016

There are two local minima points at ${p}_{1} = \left\{0 , 0\right\}$ and ${p}_{2} = \left\{1 , 0.333333\right\}$

#### Explanation:

We will searching for stationary points, qualifying then as local maxima/minima.

First we will transform the maxima/minima with inequality restrictions into an equivalent maxima/minima problem but now with equality restrictions.

To do that we will introduce the so called slack variables ${s}_{1}$ and ${s}_{2}$ such that the problem will read.

Maximize/minimize $f \left(x , y\right) = {x}^{2} + 3 x y + 9 {y}^{2}$
constrained to

{ (g_1(x,y,s_1)=x + 3 y - s_1^2=0), (g_2(x,y,s_2)=x + 3 y + s_2^2 - 2=0) :}

The lagrangian is given by

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

The condition for stationary points is

$\nabla L \left(x , y , {s}_{1} , {s}_{2} , {\lambda}_{1} , {\lambda}_{2}\right) = \vec{0}$

so we get the conditions

{ (lambda_1 + lambda_2 + 2 x + 3 y = 0), (3 lambda_1 + 3 lambda_2 + 3 x + 18 y = 0), ( -s_1^2 + x + 3 y = 0), ( -2 lambda_1 s_1 = 0), (-2 + s_2^2 + x + 3 y = 0), (2 lambda_2 s_2 = 0) :}

Solving for $\left\{x , y , {s}_{1} , {s}_{2} , {\lambda}_{1} , {\lambda}_{2}\right\}$ we have

{(x = 0., y = 0., lambda_1 = 0., s_1 = 0., lambda_2 = 0., s_2 = 1.41421), (x = 1., y = 0.333333, lambda_1 = 0., s_1 = -1.41421, lambda_2 = -3., s_2 = 0.) :}

so we have two points ${p}_{1} = \left\{0 , 0\right\}$ and ${p}_{2} = \left\{1 , 0.333333\right\}$
Point ${p}_{2}$ activates restriction ${g}_{1} \left(x , y , 0\right) = 0 , \left\{{\lambda}_{1} \ne 0 , {s}_{1} = 0\right\}$

${p}_{1}$ is qualified with $f \left(x , y\right)$

and

${p}_{2}$ is qualified with ${f}_{{g}_{2}} \left(x\right) = x \left(x - 2\right) + 4$

Computing

$\nabla f \left(0 , 0\right) = 0$

and

$\text{Eigenvalues} \left({\nabla}^{2} f \left(0 , 0\right)\right) = \left\{18.544 , 1.456\right\}$

we conclude that ${p}_{1}$ local minimum point.

Analogously for ${p}_{2}$

$\frac{d}{\mathrm{dx}} \left({f}_{{g}_{2}} \left(0\right)\right) = 0$

and

${d}^{2} / \left({\mathrm{dx}}^{2}\right) \left({f}_{{g}_{2}} \left(0\right)\right) = 2$

so ${p}_{1} , {p}_{2}$ are local minima points