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

Answer 1

Substitute #y=4/x#.

Substitute #y=4/x#. Remember that #x != 0#.
So #f# now is univariate and looks like
#f(x) = x^3 -16/x^2 - 4# graph{x^3 -16/x^2 - 4 [-10, 10, -500, 500]} Now optimize it by finding the critical points.
#f'(x) = 3x^2 + 32/x^3#
#f'(-2/root(5)3) = 0#
#x = -2/root(5)3# is a local maximum.
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Answer 2

To minimize and maximize the function ( f(x,y) = x^3 - y^2 - xy ) subject to the constraint ( xy = 4 ), we can use the method of Lagrange multipliers. The Lagrangian is given by:

[ L(x, y, \lambda) = x^3 - y^2 - xy + \lambda(xy - 4) ]

Where ( \lambda ) is the Lagrange multiplier. We need to find critical points by setting the partial derivatives of ( L ) with respect to ( x ), ( y ), and ( \lambda ) equal to zero:

  1. ( \frac{\partial L}{\partial x} = 3x^2 - y - \lambda y = 0 )
  2. ( \frac{\partial L}{\partial y} = -2y - x - \lambda x = 0 )
  3. ( \frac{\partial L}{\partial \lambda} = xy - 4 = 0 )

From equation 3, we have ( xy = 4 ). Substituting this into equations 1 and 2, we get:

  1. ( 3x^2 - y - 4\lambda = 0 ) (equation 4)
  2. ( -2y - x - 4\lambda = 0 ) (equation 5)

Multiplying equation 4 by 2 and adding it to equation 5, we get:

[ -6x^2 - 2x = 0 ]

This simplifies to ( x(x+1) = 0 ), so ( x = 0 ) or ( x = -1 ). If ( x = 0 ), then ( y = \pm 2 ). If ( x = -1 ), then ( y = \mp 4 ).

Thus, the critical points are ( (0, 2) ), ( (0, -2) ), ( (-1, -4) ), and ( (-1, 4) ). To determine which of these points correspond to the minimum or maximum of ( f ), we evaluate ( f ) at each point:

  1. ( f(0, 2) = 0^3 - 2^2 - 0\cdot2 = -4 )
  2. ( f(0, -2) = 0^3 - (-2)^2 - 0\cdot(-2) = -4 )
  3. ( f(-1, -4) = (-1)^3 - (-4)^2 - (-1)\cdot(-4) = 3 + 16 + 4 = 23 )
  4. ( f(-1, 4) = (-1)^3 - 4^2 - (-1)\cdot4 = -1 - 16 + 4 = -13 )

Therefore, the minimum value of ( f ) is -4 and occurs at ( (0, 2) ) and ( (0, -2) ), while the maximum value is 23 and occurs at ( (-1, -4) ).

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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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