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How do you minimize and maximize #f(x,y)=2x^2-x/(2x-3y)# constrained to #1<yx^2+xy^2<16#?

Answer 1

Emilia Aguilar

See below.

With the Lagrange Multipliers technique we get the following results

#((f,x,y),(0,0,-128),(31.7141,-4,2),(0,0.01248,-8.04377),(4.75284,-1.58758,0.779102))#

Attached the feasible region and some maxima/minima points

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Answer 2

Axel Alvarez

To minimize and maximize the function ( f(x, y) = \frac{{2x^2 - x}}{{2x - 3y}} ) constrained to ( 1 < yx^2 + xy^2 < 16 ), follow these steps:

Find critical points by solving the system of equations formed by setting the partial derivatives of ( f(x, y) ) equal to zero and solving for ( x ) and ( y ).
Evaluate ( f(x, y) ) at each critical point.
Evaluate ( f(x, y) ) at the boundary points of the constraint region ( 1 < yx^2 + xy^2 < 16 ).
Compare all the values obtained in steps 2 and 3 to determine the minimum and maximum values of ( f(x, y) ).

This approach will yield the minimum and maximum values of ( f(x, y) ) 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.

How do you minimize and maximize #f(x,y)=2x^2-x/(2x-3y)# constrained to #1&lt;yx^2+xy^2&lt;16#?

How do you minimize and maximize #f(x,y)=2x^2-x/(2x-3y)# constrained to #1<yx^2+xy^2<16#?