# How do you minimize and maximize f(x,y)=xsqrt(xy+y) constrained to 0<xy-y^2<5?

Jun 5, 2016

${p}_{1} = \left\{- 4.69709 , - 1.63044\right\}$ maximum and ${p}_{2} = \left\{4.78125 , 1.54499\right\}$ minimum.

#### 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 $f \left(x , y\right) = x \sqrt{x y + y}$
constrained to

{ (g_1(x,y,s_1)=x y - y^2 - s^2=0), (g_2(x,y,s_2)=x y - y^2 + s_2^2 - 5=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 y + lambda_2 y + (x y)/(2 sqrt[y + x y]) + sqrt[y + x y] = 0), (lambda_1 (x - 2 y) + lambda_2 (x - 2 y) + (x (1 + x))/(2 sqrt[y + x y]) = 0), ( -s_1^2 + x y - y^2 = 0), (-2 lambda_1 s_1 = 0), (-5 + s_2^2 + x y - 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 = -4.69709, y = -1.63044, lambda_1 = 0., s_1 = -2.23607,lambda_2 = 2.4624,s_2 = 0.), (x = 4.78125, y = 1.54499, lambda_1 = 0., s_1 = -2.23607, lambda_2 = -2.73431, s_2 = 0.) :}

so we have two points

${p}_{1} = \left\{- 4.69709 , - 1.63044\right\}$

and

${p}_{2} = \left\{4.78125 , 1.54499\right\}$

Point ${p}_{1}$ activates restriction ${g}_{2} \left(x , y , 0\right) = 0 , \left\{{\lambda}_{2} \ne 0 , {s}_{2} = 0\right\}$

Point ${p}_{2}$ activates restriction ${g}_{2} \left(x , y , 0\right) = 0 , \left\{{\lambda}_{2} \ne 0 , {s}_{2} = 0\right\}$

${p}_{1}$ is qualified with ${f}_{{g}_{2}} \left(x\right) = \frac{x \sqrt{\left(1 + x\right) \left(x + \sqrt{{x}^{2} - 20}\right)}}{\sqrt{2}}$

and

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

Computing

$\frac{d}{\mathrm{dx}} \left({f}_{{g}_{2}} \left(- 4.69709\right)\right) = 0$

and

${d}^{2} / \left({\mathrm{dx}}^{2}\right) \left({f}_{{g}_{2}} \left(- 4.69709\right)\right) = - 8.19783$

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

Analogously for ${p}_{2}$

$\frac{d}{\mathrm{dx}} \left({f}_{{g}_{2}} \left(0\right)\right) = 4.78125$

and

${d}^{2} / \left({\mathrm{dx}}^{2}\right) \left({f}_{{g}_{2}} \left(4.78125\right)\right) = 6.22258$

so ${p}_{1} , {p}_{2}$ are local maximum and minimum points