# How do you minimize and maximize #f(x,y)=x^2+y^3# constrained to #0<x+3xy<4#?

A local minimum at

a local minimum also at

Introducing the so called slack variables

Find local minima, maxima of

subject to

The lagrangian is

The determination of stationary points is done solving for

or

Solving we get

and

Both points are at the boundaries of

Their qualification must be done with

Attached a figure with a contour mapping with the found points

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To minimize and maximize ( f(x, y) = x^2 + y^3 ) constrained to ( 0 < x + 3xy < 4 ), we use the method of Lagrange multipliers. The Lagrangian function is:

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

We then solve the following system of equations:

[ \frac{\partial L}{\partial x} = 2x + 3y\lambda = 0 ] [ \frac{\partial L}{\partial y} = 3y^2 + 3x\lambda = 0 ] [ \frac{\partial L}{\partial \lambda} = x + 3xy - 4 = 0 ]

Solving these equations simultaneously will give us the critical points. We then evaluate ( f(x, y) ) at these points and check the boundaries of the constraint for possible extrema.

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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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