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

Answer 1

#p_{max} = {-0.0742241,6.32052,0.}# maximum point of local maxima.

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) =(2 x - 3 y)^2 - 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(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^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 #{x,y,s_1,s_2,lambda_1,lambda_2}# 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(x,y,0) = 0# and the last 5 activate restriction #g_2(x,y,s_2) = 0#

The first 5 points must be evaluated on

#f_{g_1}(x,y) = (2 + 18 x - 5 x^4 pm 3 x^(5/2) sqrt[x^3-4])/(2 x^2)#

and the last 5 on

#f_{g_2}(x,y)=(2 + 54 x - 5 x^4 pm 3 x^(5/2) sqrt[x^3-12])/(2 x^2)#

After qualifying we obtain

#p_{max} = {-0.0742241,6.32052,0.}# maximum point of local maxima.

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Answer 2

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

  1. First, calculate the gradient of the objective function and the constraint function.
  2. Set up the Lagrangian by adding the product of Lagrange multipliers and the constraint function to the objective function.
  3. Find the critical points of the Lagrangian by setting its partial derivatives with respect to ( x ), ( y ), and the Lagrange multipliers equal to zero.
  4. Solve the resulting system of equations to find the critical points.
  5. Evaluate the objective function at the critical points.
  6. The minimum and maximum values of ( f(x, y) ) will correspond to the smallest and largest values obtained from step 5.

This method will yield the minimum and maximum values of ( f(x, y) ) subject to 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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