Question

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

Answer 1

The function:

`f(x,y) = (xy)/((x-2)(y-4))`

constrained by `xy=2`

has two local maximums for `(x,y) = (-1,-2)` and `(x,y) = (1,2)`

Explanation:

From the constraint equation:

`xy = 2`

we can derive:

`y=2/x`

and substitute it in the original equation:

`g(x) = f(x,y(x)) = 2/((x-2)(2/x-4)) = (2x)/((x-2)(2-4x))`

We can now find the local extrema for `g(x)`:

`g(x) = (2x)/( 2x-4-4x^2+8x) = - (x)/(2x^2-5x+2)`

`(dg)/dx = - ( 2x^2-5x+2 - x(4x-5))/(2x^2-5x+2)`

`(dg)/dx = - ( 2x^2-5x+2 - 4x^2+5x)/(2x^2-5x+2)`

`(dg)/dx = ( 2x^2-2)/(2x^2-5x+2) = (2(x^2-1))/((x-2)(2x-1))`

The critical points are determined by the equation:

`(dg)/dx = 0 => x=+-1`

Looking at the sign of the derivative we can see that:

`(dg)/dx > 0` for `x<-1`

`(dg)/dx < 0` for `-1 < x< 1/2`

So that `x=-1` is a local maximum and:

`(dg)/dx > 0` for `1/2 < x < 1`

`(dg)/dx < 0` for `1 < x < 2`

so that also `x=1` is a local maximum.