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

Feb 1, 2017

There is one extrema at $\left(3 , 3 , 27\right)$

Explanation:

We have:

$f \left(x , y\right) = x y + \frac{27}{x} + \frac{27}{y}$

And so we derive the partial derivatives:

$\frac{\partial f}{\partial x} = y - \frac{27}{x} ^ 2 \setminus \setminus \setminus$ and $\setminus \setminus \setminus \frac{\partial f}{\partial y} = x - \frac{27}{y} ^ 2$

At an extrema or saddle points we have:

$\frac{\partial f}{\partial x} = 0 \setminus \setminus \setminus$ and $\setminus \setminus \setminus \frac{\partial f}{\partial y} = 0 \setminus \setminus \setminus$ simultaneously:

i.e. a simultaneous solution of:

$y - \frac{27}{x} ^ 2 = 0 \implies {x}^{2} y = 27$
$x - \frac{27}{y} ^ 2 = 0 \implies x {y}^{2} = 27$

Subtracting these equations gives:

$\setminus \setminus \setminus \setminus {x}^{2} y - x {y}^{2} = 0$
$\therefore x y \left(x - y\right) = 0$
 :. 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:

${x}^{3} = 27 \implies x = y = 3$

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

$f \left(3 , 3\right) = 9 + 9 + 9 = 27$

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) 