How do you maximize and minimize #f(x,y)=x-xy^2# constrained to #0<=x^2+y<=1#?

Answer 1

There are several local maxima/minima

With the so called slack variables #s_1, s_2# we transform the maximization/minimization with inequality constraints problem, into a formulation amenable for the Lagrange Multipliers technique.

Now the lagrangian formulation reads:

minimize/maximize
#f(x,y) =x - x y^2#

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

forming the lagrangian

# 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 local minima/maxima points are included into the lagrangian stationary points found by solving

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

or

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

Solving for #x,y,s_1,s_2,lambda_1,lambda_2# we obtain

#( (x = 0, y=1., lambda_1 = 0, s_1 = -1., lambda_2= 0, s_2 = 0), (x = -1.09545, y= -0.2, lambda_1 = 0, s_1= -1., lambda_2 = 0.438178, s_2 = 0), (x = 1.09545, y= -0.2, lambda_1= 0, s_1= -1., lambda_2 = -0.438178, s_2 = 0), (x = -0.66874, y= -0.447214, lambda_1 = 0.59814, s_1 =0, lambda_2 = 0, s_2= -1.), (x = 0.66874, y= -0.447214, lambda_1 = -0.59814, s_1 = 0, lambda_2 = 0, s_2 = -1.) )#

Those five points must be qualified. The first second and thirt activate constraint #g_2(x,y,0)# and the two other activate constraint #g_1(x,y,0)# Their qualification will be done with

#f@g_1 (x) =x - x^5# and
#f@g_2(x) =-x^3 (x^2-2)#

giving

#d^2/(dx^2)f@g_1 ( -0.66874) =5.9814# local minimum
#d^2/(dx^2)f@g_1 (0.66874) = 5.90567# local maximum
#d^2/(dx^2)f@g_2 (0) = 0# not decidable
#d^2/(dx^2)f@g_2 (-1.09545) = 13.1453# local minimum
#d^2/(dx^2)f@g_2 ( 1.09545) =-13.1453# local maximum

Attached a figure with the #f(x,y)# contour map inside the feasible region, with the local maxima/minima points.

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Answer 2

To maximize and minimize ( f(x,y) = x - xy^2 ) constrained to ( 0 \leq x^2 + y \leq 1 ), you can use the method of Lagrange multipliers. Here are the steps:

  1. Set up the Lagrangian function: [ L(x, y, \lambda) = x - xy^2 + \lambda(x^2 + y - 1) ]

  2. Compute the partial derivatives of ( L ) with respect to ( x ), ( y ), and ( \lambda ), and set them equal to zero: [ \frac{\partial L}{\partial x} = 1 - y^2 + 2\lambda x = 0 ] [ \frac{\partial L}{\partial y} = -2xy + \lambda = 0 ] [ \frac{\partial L}{\partial \lambda} = x^2 + y - 1 = 0 ]

  3. Solve the system of equations obtained in step 2 to find the critical points.

  4. Check the points obtained in step 3 to determine which points satisfy the given constraint ( 0 \leq x^2 + y \leq 1 ).

  5. Evaluate ( f(x, y) ) at each critical point satisfying the constraint to find the maximum and minimum values.

  6. Compare the values obtained to find the maximum and minimum 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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