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

1 Answer
Feb 1, 2017

Answer:

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)

enter image source here