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

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Answer 1 · 2017-02-01T13:57:14

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

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)

enter image source here

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