Question

Find the maximum and minimum values of the function f(x,y)= x^2 - 4xy + 2y^2 + 3x. I've calculated the first and second derivatives and found Hf(P) but i don't know what to do after that

Answer 1

you have used partial derivatives right?

Answer 2

the xy part tells me its a conic thats been rotated at the origin

Answer 3

its elliptic if i see it right

Answer 4

hyperbolic paraboloid .. i was close :)

Answer 5

its only going to have a low spot i believe, the graph is a pretty much a piece of paper lifted at the corners

Answer 6

find your partials at zero like you would any other "curve"

Answer 7

the partial derivatives give you the equation for the normal at any given point

Answer 8

\begin{eqnarray*}
\frac{\partial f}{\partial x}&=&2x-4y+3 \\
\frac{\partial f}{\partial y}&=&-4x+4y
\end{eqnarray*}
Both of the above are zero at stationary values.

Answer 9

I rule itself is foggy; fx(x,y) + fy(x,y) ...something fxy(x,y)

Answer 10

Fxx*Fyy-(Fxy)^2 rules perhaps?

Answer 11

he'll then need to find \[\frac{\partial^2f}{\partial x^2},\frac{\partial^2f}{\partial y^2}, \frac{\partial^2f}{\partial x\partial y}\]
then evaluate
\[\Delta = \Big(\frac{\partial^2f}{\partial x\partial y}\Big)^2-\Big(\frac{\partial^2f}{\partial x^2}\Big)\Big(\frac{\partial^2f}{\partial x^2}\Big)\]
A point is maximum if
\[\Delta<0 , \frac{\partial^2f}{\partial x^2}<0\]
A point is minimum if
\[\Delta<0 , \frac{\partial^2f}{\partial x^2}>0\]
A point is saddle point if
\[\Delta>0\]

Answer 12

Fancy typing always makes it better :) those are them