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

Answer 1

Twho local maxima at #{0.5,0.5}# and #{1.5,1.5}#

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(x,y)=(e^(yx)-e^(-yx))/(2yx)#

subjected to
#g_1(x,y,s_1) = x^2/y+y^2/x-s_1^2-1=0#
#g_2(x,y,s_2) = x^2/y+y^2/x-s_2^2-3=0#

The lagrangian reads
#L(x,y,s_1,s_2,lambda_1,lambda_2)=f(x,y)+lambda_1g_1(x,y,s_1)+lambda_2 g_2(x.y,s_2)#

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

The stationary points are the solutions of
#grad L(x,y,s_1,s_2,lambda_1,lambda_2)=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 #{1.5,1.5}# activates #g_2(x,y,0)# and the second #{0.5,0.5}# activates #g_1(x,y,0)# 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
#(d^2)/(dx^2)f@g_1(0.5,0.5)=-1.5094# local maximum
#(d^2)/(dx^2)f@g_2(1.5,1.5)=-21.6928# local maximum

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

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Answer 2

To minimize and maximize ( f(x,y) = \frac{e^{yx} - e^{-yx}}{2yx} ) subject to ( 1 < \frac{x^2}{y} + \frac{y^2}{x} < 3 ), we can use the method of Lagrange multipliers.

Define the Lagrangian function ( L ) as:

[ L(x, y, \lambda) = \frac{e^{yx} - e^{-yx}}{2yx} + \lambda \left( \frac{x^2}{y} + \frac{y^2}{x} - 3 \right) ]

Taking partial derivatives with respect to ( x ), ( y ), and ( \lambda ) and setting them to zero will give us the critical points.

  1. Partial derivative with respect to ( x ):

[ \frac{\partial L}{\partial x} = \frac{y(e^{yx} + e^{-yx})}{2yx^2} + \lambda \left( \frac{2x}{y} - \frac{y^2}{x^2} \right) = 0 ]

  1. Partial derivative with respect to ( y ):

[ \frac{\partial L}{\partial y} = \frac{x(e^{yx} - e^{-yx})}{2y^2x} + \lambda \left( \frac{x^2}{y^2} + \frac{2y}{x} \right) = 0 ]

  1. Partial derivative with respect to ( \lambda ):

[ \frac{\partial L}{\partial \lambda} = \frac{x^2}{y} + \frac{y^2}{x} - 3 = 0 ]

Solve this system of equations to find the critical points of ( f(x,y) ). After finding these critical points, evaluate ( f(x,y) ) at each point and compare the values to determine the minimum and maximum values of ( f(x,y) ) within the given constraint.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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