# Thread: stationary point on a surface

1. ## stationary point on a surface

$z= 3x^2 + 7x + 2y^2 - 5y -xy + 13$
Find the stationary point and determine the nature of the point(g).
My question is did any step im do wrongly?

STATIONARY POINT
$\frac{\partial z}{\partial x} = 6x + 7 - y$

$\frac{\partial z}{\partial y} = 4y - 5 - x$
$\frac{\partial z}{\partial x} = 0 ,\frac{\partial z}{\partial y} = 0$
$6x + 7 - y = 0$
$4y - 5 - x = 0$
Then x and y value is:
$x = \frac{-3}{19}$
$y = \frac{23}{19}$
Substitute x and y into original equation(z), then get $z=\frac{3263}{361}$

stationary point is at ( $\frac{-3}{19}$, $\frac{23}{19}$, $\frac{3263}{361}$)

Now i want find the nature of the point(g):
$g = (\frac{\partial^2 z}{\partial x^2})(\frac{\partial^2 z}{\partial y^2}) -(\frac{\partial^2 z}{\partial x\partial y} )^2$
$= 6(4)-(-1)^2$
$=23>0$
Stationary point is a minimum point

2. Originally Posted by wkn0524
$z= 3x^2 + 7x + 2y^2 - 5y -xy + 13$
Find the stationary point and determine the nature of the point(g).
My question is did any step im do wrongly?

STATIONARY POINT
$\frac{\partial z}{\partial x} = 6x + 7 - y$

$\frac{\partial z}{\partial y} = 4y - 5 - x$
$\frac{\partial z}{\partial x} = 0 ,\frac{\partial z}{\partial y} = 0$
$6x + 7 - y = 0$
$4y - 5 - x = 0$
Okay so far!

Then x and y value is:
$x = \frac{-3}{19}$
$y = \frac{23}{19}$
Not even close! Solve those equations again!

Substitute x and y into original equation(z), then get $z=\frac{3263}{361}$

stationary point is at ( $\frac{-3}{19}$, $\frac{23}{19}$, $\frac{3263}{361}$)

Now i want find the nature of the point(g):
$g = (\frac{\partial^2 z}{\partial x^2})(\frac{\partial^2 z}{\partial y^2}) -(\frac{\partial^2 z}{\partial x\partial y} )^2$
$= 6(4)-(-1)^2$
$=23>0$
Stationary point is a minimum point

3. Ouch, i knew my mistakes...i did a wrong common factor for x and y.

$x = -1$
$y = 1$

Thanks ya