How do you minimize and maximize #f(x,y)=(x-y)/(x^3-y^3# constrained to #2<xy<4#?

Answer 1

See below.

How do you minimize and maximize #f(x,y)=(x-y)/(x^3-y^3)# constrained to #2 < xy <4 #?

Considering that #(x^3-y^3)/(x-y)=x^2+ xy+y^2# we have now the optimization problem

Find extrema of

#f(x,y) =1/(x^2+x y + y^2)#

subjected to

#g_1(x,y,s_1) = x y - 2 - s_1^2 =0#
#g_2(x,y,s_2) = x y - 4 + s_2^2=0#

#s_1,s_2# are known as slack variables. They are used to transform the inequality into equality relations.

The lagrangian is

#L(x,y,s_1,s_2,lambda_1,lambda_2)=f(x,y)+lambda_1g_1(x,y,s_1)+lambda_2g_2(x,y,s_2)# and it's stationary points are the points obeying

#grad L = vec 0# or

#{ (lambda_1 y + lambda_2 y - (2 x + y)/(x^2 + x y + y^2)^2 = 0), (lambda_1 x + lambda_2 x - (x + 2 y)/(x^2 + x y + y^2)^2 = 0), (-2 lambda_1 s_1 = 0), (2 lambda_2 s_2 = 0), (s_1^2 + x y - 2= 0), (s_2^2 + x y - 4= 0):}#

Solving for #x,y,s_1,s_2,lambda_1,lambda_2# we obtain
#p_1,p_2,p_3,p_4#

#( (x = -2., y = -2., s_1 = -1.41421, s_2 = 0, lambda_1 = 0, lambda_2 -> 0.0208333), (x = 2., y= 2., s_1 = -1.41421, s_2 = 0, lambda_1= 0, lambda_2 = 0.0208333), (x = -1.41421, y= -1.41421, s_1 = 0, s_2 = -1.41421, lambda_1 =0.0833333, lambda_2 = 0), (x = 1.41421, y = 1.41421, s_1 = 0, s_2= -1.41421, lambda_1= 0.0833333, lambda_2 = 0))#

The stationary points qualification is done according to the sign of

#(d^2(f@g_1)(s_1=0))/(dx^2)# and
#(d^2(f@g_2)(s_2=0))/(dx^2)#

Here

#(f@g_1)(x) = 1/(2 + 4/x^2 + x^2)#
#(f@g_2)(x) = 1/(4 + 16/x^2 + x^2)#

and

#(d^2(f@g_1)(s_1=0))/(dx^2) < 0# and
#(d^2(f@g_2)(s_2=0))/(dx^2) < 0# for all points so

#p_1,p_2,p_3,p_4# are all local maxima.

Attached a plot showing the feasible region and the points location.

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Answer 2

To minimize and maximize the function ( f(x, y) = \frac{x - y}{x^3 - y^3} ) subject to the constraint ( 2 < xy < 4 ), you can follow these steps:

  1. Find the critical points of ( f(x, y) ) within the given constraint by setting the partial derivatives of ( f(x, y) ) with respect to ( x ) and ( y ) equal to zero and solving for ( x ) and ( y ).
  2. Evaluate the function ( f(x, y) ) at these critical points as well as at the endpoints of the constraint region (( xy = 2 ) and ( xy = 4 )).
  3. Compare the values obtained to determine the minimum and maximum values of ( f(x, y) ) within the given constraint.

These steps will help you find the minimum and maximum values of ( f(x, y) ) subject to the constraint ( 2 < xy < 4 ).

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Answer 3

To minimize and maximize the function ( f(x,y) = \frac{x-y}{x^3-y^3} ) constrained to ( 2 < xy < 4 ), we first find the critical points by taking the partial derivatives of ( f(x, y) ) with respect to ( x ) and ( y ), setting them equal to zero, and then solving for ( x ) and ( y ). After obtaining the critical points, we evaluate the function at these points as well as at the boundary points of the constraint region ( 2 < xy < 4 ). Finally, we compare the values of the function at these points to determine the minimum and maximum values.

After finding the critical points, we also need to check the boundary points of the constraint region ( 2 < xy < 4 ). This involves evaluating the function at points where ( xy = 2 ) and ( xy = 4 ), and comparing these values with the values at the critical points.

We then select the smallest and largest values among the values obtained at the critical points and the boundary points to determine the minimum and maximum values of the function within the given constraint region.

To summarize:

  1. Find critical points by taking partial derivatives of ( f(x, y) ) with respect to ( x ) and ( y ), setting them equal to zero, and solving for ( x ) and ( y ).
  2. Evaluate the function at the critical points.
  3. Evaluate the function at the boundary points of the constraint region ( 2 < xy < 4 ).
  4. Compare all values obtained in steps 2 and 3 to determine the minimum and maximum values of the function within the given 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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