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

Two local maxima at

Two local minima at

With the so called slack variables

Now the lagrangian formulation reads:

minimize/maximize

subjected to

forming the lagrangian

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

or

Solving for

Those eight points are truly four. The first and second activate constraint

giving

Attached a figure with the

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

First, you set up the Lagrangian:

( L(x, y, \lambda) = \frac{{2x-y}}{{x}} + \frac{{x-2y}}{{x^2}} + \lambda(yx^2 + xy^2 - 1) - \mu(yx^2 + xy^2 - 3) )

Then, take the partial derivatives with respect to ( x ), ( y ), and ( \lambda ), and set them equal to zero:

( \frac{{\partial L}}{{\partial x}} = 0 ), ( \frac{{\partial L}}{{\partial y}} = 0 ), ( \frac{{\partial L}}{{\partial \lambda}} = 0 ).

Solve the resulting system of equations to find the critical points. Then, evaluate the function at these critical points and the boundary points of the constraint ( 1 < yx^2 + xy^2 < 3 ) to find the minimum and maximum values of ( f(x, y) ).

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To minimize and maximize the function ( f(x, y) = (2x - y) + \frac{x - 2y}{x^2} ) constrained to ( 1 < yx^2 + xy^2 < 3 ), follow these steps:

- Find the critical points of the function by taking the partial derivatives with respect to ( x ) and ( y ) and setting them equal to zero.
- Use the second derivative test to determine whether the critical points are maxima, minima, or saddle points.
- Evaluate the function at the critical points and at the boundary points of the constraint region.
- Compare the values obtained to find the maximum and minimum values of the function within the constraint region.

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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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