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

See below.

There is a local minimum at

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To minimize and maximize ( f(x,y) = ye^{2x} - \ln\left(\frac{y}{x}\right) ) constrained to ( 0 < xy - y + x < 1 ), we need to use the method of Lagrange multipliers. First, we form the Lagrangian:

[ L(x, y, \lambda) = ye^{2x} - \ln\left(\frac{y}{x}\right) + \lambda(xy - y + x - 1) ]

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

[ \frac{\partial L}{\partial x} = 2ye^{2x} - \frac{\lambda y}{x} + \lambda = 0 ] [ \frac{\partial L}{\partial y} = e^{2x} - \frac{1}{y} + \lambda(x - 1) = 0 ] [ \frac{\partial L}{\partial \lambda} = xy - y + x - 1 = 0 ]

Solving these equations simultaneously will give the critical points. After obtaining the critical points, evaluate the function ( f(x, y) ) at each critical point and the boundary points ( xy - y + x = 0 ) and ( xy - y + x = 1 ). Then compare the values to find the minimum and maximum values of ( f(x, y) ).

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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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