# stationary point on a surface

#### wkn0524

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

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

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

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

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

Last edited:

#### HallsofIvy

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

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

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

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

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

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

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

• wkn0524

#### wkn0524

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

$$\displaystyle x = -1$$
$$\displaystyle y = 1$$

Thanks ya(Clapping)