stationary point on a surface

Mar 2010
26
0
Malaysia
\(\displaystyle z= 3x^2 + 7x + 2y^2 - 5y -xy + 13\)
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
Apr 2005
20,249
7,909
\(\displaystyle z= 3x^2 + 7x + 2y^2 - 5y -xy + 13\)
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
 
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Mar 2010
26
0
Malaysia
Ouch, i knew my mistakes...i did a wrong common factor for x and y.

\(\displaystyle x = -1\)
\(\displaystyle y = 1\)

Thanks ya(Clapping)