How do you minimize and maximize f(x,y)=(2x-3y)^2-1/x^2 constrained to 1<yx^2+xy^2<3?

Jun 6, 2016

${p}_{\max} = \left\{- 0.0742241 , 6.32052 , 0.\right\}$ maximum point of local maxima.

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 $f \left(x , y\right) = {\left(2 x - 3 y\right)}^{2} - \frac{1}{x} ^ 2$
constrained to

{ (g_1(x,y,s_1)=x^2 y - x y^2 - s_1^2 - 1=0), (g_2(x,y,s_2)=x^2 y - x y^2 + s_2^2 - 3=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

{ (2/x^3 + 4 (2 x - 3 y) + lambda_1 (2 x y - y^2) + lambda_2 (2 x y - y^2) = 0), (-6 (2 x - 3 y) + lambda_1 (x^2 - 2 x y) + lambda_2 (x^2 - 2 x y) = 0), ( -1 - s_1^2 + x^2 y - x y^2 = 0), (-2 lambda_1 s_1 = 0), (-3 + s_2^2 + x^2 y - x y^2 = 0), (2 lambda_2 s_2 = 0) :}

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

{ (x = -1.47669, y= -1.84394, lambda_1= -4.73797, s_1 = 0., lambda_2 = 0., s_2= -1.41421), (x = -0.964714, y= 0.64425, lambda_1= -10.6606, s_1= 0., lambda_2= 0., s_2 = -1.41421), (x= -0.227896, y= 1.9839, lambda_1 = -40.2068, s_1 = 0., lambda_2 = 0., s_2= -1.41421), (x= -0.219711, y= -2.24609, lambda_1= -40.2608, s_1 = 0., lambda_2= 0., s_2 = -1.41421), (x = 1.64519, y = 1.08494, lambda_1 = -0.247151, s_1= 0., lambda_2= 0., s_2= -1.41421), (x = -2.22844, y = -2.72286, lambda_1 = 0., s_1 = -1.41421, lambda_2 = -3.10624, s_2 =0.), (x = -1.49657, y= 0.853125, lambda_1= 0., s_1 = -1.41421, lambda_2= -6.95042, s_2 = 0.), (x= -0.0742241, y = 6.32052, lambda_1 = 0., s_1= -1.41421, lambda_2= -121.49, s_2 = 0.), (x = -0.0739366, y = -6.40695, lambda_1 = 0., s_1 = -1.41421,lambda_2 = -121.491,s_2=0.), (x = 2.37908, y= 1.58197, lambda_1 = 0., s_1= -1.41421, lambda_2 = -0.0392951, s_2 = 0.) :}

so we have 10 points which are potential local maxima/minima points. The first 5 points activate ${g}_{1} \left(x , y , 0\right) = 0$ and the last 5 activate restriction ${g}_{2} \left(x , y , {s}_{2}\right) = 0$

The first 5 points must be evaluated on

${f}_{{g}_{1}} \left(x , y\right) = \frac{2 + 18 x - 5 {x}^{4} \pm 3 {x}^{\frac{5}{2}} \sqrt{{x}^{3} - 4}}{2 {x}^{2}}$

and the last 5 on

${f}_{{g}_{2}} \left(x , y\right) = \frac{2 + 54 x - 5 {x}^{4} \pm 3 {x}^{\frac{5}{2}} \sqrt{{x}^{3} - 12}}{2 {x}^{2}}$

After qualifying we obtain

${p}_{\max} = \left\{- 0.0742241 , 6.32052 , 0.\right\}$ maximum point of local maxima.