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

Jul 1, 2016

Twho local maxima at $\left\{0.5 , 0.5\right\}$ and $\left\{1.5 , 1.5\right\}$

#### Explanation:

Completing with the so called slack variables ${s}_{1} , {s}_{2}$ we propose an equivalent problem to be handled with the Lagrangian Multipliers technique.

Minimize/maximize
$f \left(x , y\right) = \frac{{e}^{y x} - {e}^{- y x}}{2 y x}$

subjected to
${g}_{1} \left(x , y , {s}_{1}\right) = {x}^{2} / y + {y}^{2} / x - {s}_{1}^{2} - 1 = 0$
${g}_{2} \left(x , y , {s}_{2}\right) = {x}^{2} / y + {y}^{2} / x - {s}_{2}^{2} - 3 = 0$

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

The lagrangian stationary points includes the local maxima/minima points.

The stationary points are the solutions of
$\nabla L \left(x , y , {s}_{1} , {s}_{2} , {\lambda}_{1} , {\lambda}_{2}\right) = \vec{0}$

or

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

Solving for $x , y , {s}_{1} , {s}_{2} , {\lambda}_{1} , {\lambda}_{2}$ with an iterative procedure such as Newton-Raphson's we obtain

( (x = 1.5, y = 1.5, lambda_1 = 0, s_1=-1.41421, lambda_2 = -1.80774, s_2 = 0), (x= 1.5, y = 1.5, lambda_1 = 0, s_1 = 1.41421, lambda_2= -1.80774, s_2 = 0), (x = 0.5, y = 0.5, lambda_1 = -0.0419277, s_1 = 0, lambda_2 = 0, s2 = -1.41421), (x = 0.5, y=0.5, lambda_1= -0.0419277, s_1= 0, lambda_2= 0, s_2= 1.41421) )

There are four solutions but for two points. The first $\left\{1.5 , 1.5\right\}$ activates ${g}_{2} \left(x , y , 0\right)$ and the second $\left\{0.5 , 0.5\right\}$ activates ${g}_{1} \left(x , y , 0\right)$ so their qualification must be against

f@g_1 =((2/3)^(1/3) x)/(-9 x^3 + sqrt[3] sqrt[-4 x^3 + 27 x^6])^( 1/3) + (-9 x^3 + sqrt[3] sqrt[-4 x^3 + 27 x^6])^(1/3)/( 2^(1/3) 3^(2/3))
and
f@g_2=(3 2^(1/3) x)/(-x^3 + sqrt[-108 x^3 + x^6])^( 1/3) + (-x^3 + sqrt[-108 x^3 + x^6])^(1/3)/2^(1/3)

Calculating
$\frac{{d}^{2}}{{\mathrm{dx}}^{2}} f \circ {g}_{1} \left(0.5 , 0.5\right) = - 1.5094$ local maximum
$\frac{{d}^{2}}{{\mathrm{dx}}^{2}} f \circ {g}_{2} \left(1.5 , 1.5\right) = - 21.6928$ local maximum

Attached the figure with the feasible region showing the $f \left(x , y\right)$ contour map and the local extrema points.