Question

What are the extrema and saddle points of `f(x, y) = xy+27/x+27/y`?

Answer 1

There is one extrema at `(3,3,27)`

Explanation:

We have:

`f(x,y) = xy + 27/x + 27/y`

And so we derive the partial derivatives:

`(partial f) / (partial x) = y - 27/x^2 \ \ \` and `\ \ \ (partial f) / (partial y) = x - 27/y^2`

At an extrema or saddle points we have:

`(partial f) / (partial x) = 0 \ \ \` and `\ \ \ (partial f) / (partial y) = 0 \ \ \` simultaneously:

i.e. a simultaneous solution of:

`y - 27/x^2 = 0 => x^2y = 27`
`x - 27/y^2 = 0 => xy^2 = 27`

Subtracting these equations gives:

`\ \ \ \ x^2y-xy^2 = 0`
`:. xy(x-y) = 0`
`:. x=0; y=0; x=y`

We can eliminate `x=0; y=0` and so `x=y` is the only valid solution, which leads to:

`x^3 = 27 => x=y=3`

And with `x=y=3`, we have:

`f(3,3) = 9+9+9=27`

Hence there is only one critical point which occurs at (3,3,27) which can be seen on this plot (which includes the tangent plane)