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

Answer 1

The points #{x = pm 5.15905, y = pm 3.86559}# are local minima

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(x,y) = (x - y) (x + y) + sqrt[x y]#
constrained to

#{ (g_1(x,y,s_1)=x y - y^2 - s_1^2=0), (g_2(x,y,s_2)=x y - y^2 + s_2^2 - 5=0) :}#

The lagrangian is given by

#L(x,y,s_1,s_2,lambda_1,lambda_2) = f(x,y)+lambda_1 g_1(x,y,s_1)+lambda_2g_2(x,y,s_2)#

The condition for stationary points is

#grad L(x,y,s_1,s_2,lambda_1,lambda_2)=vec 0#

so we get the conditions

#{ (2 x + lambda_1 y + lambda_2 y + y/(2 sqrt[x y]) = 0), (lambda_1 (x - 2 y) + llambda_2 (x - 2 y) - 2 y + x/(2 sqrt[x y]) = 0), (-s_1^2 + x y - y^2 = 0), (-2 lambda_1 s_1 = 0), (-5 + s2^2 + x y - y^2 = 0), (2 lambda_2 s_2 = 0) :}#

Solving for #{x,y,s_1,s_2,lambda_1,lambda_2}# we have

#{(x= -5.15905, y= -3.86559, lambda_1 = 0., s_1 = -2.23607, lambda_2= -2.78118, s_2= 0.), (x = 5.15905, y = 3.86559, lambda_1= 0., s_1 = -2.23607, lambda_2= -2.78118, s_2= 0.) :}#

The restriction #g_2(x,y)# is active (#lambda_2 ne 0# and #s_2=0#) so for qualifying the maxima/minima we produce

#f_{g_2}(x) = 1/2 (10 + sqrt[2] sqrt[x (x - sqrt[-20 + x^2])] pm x (x + sqrt[x^2-20]))#

Computing

#d/(dx)(f_{g_2}( pm5.15905)) = 0#

and

#d^2/(dx^2)(f_{g_2}( pm5.15905)) = 1.64412#

we conclude that the found solutions are local minima points.

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Answer 2

To minimize and maximize the function ( f(x, y) = (x - y)(x + y) + \sqrt{xy} ) constrained to ( 0 < xy - y^2 < 5 ), follow these steps:

  1. Find critical points of ( f(x, y) ) by setting partial derivatives with respect to ( x ) and ( y ) equal to zero.
  2. Evaluate the function at critical points and boundary points of the constraint region.
  3. Compare the values obtained in step 2 to determine the minimum and maximum values of ( f(x, y) ) within the constraint region.

The critical points can be found by solving the system of equations:

[ \frac{\partial f}{\partial x} = 2x - 2y + \frac{1}{2\sqrt{xy}}(y + 2x) = 0 ] [ \frac{\partial f}{\partial y} = -2x + 2y + \frac{1}{2\sqrt{xy}}(x - 2y) = 0 ]

Solve these equations simultaneously to find the critical points. Then evaluate ( f(x, y) ) at these points and at the boundary points of the constraint region ( 0 < xy - y^2 < 5 ). Compare the values obtained to determine the minimum and maximum values of ( f(x, y) ) within the constraint region.

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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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