Question

Find the critical points of the given function and then determine whether it is a local maximum, local minimum, or saddle point. f(x, y) = 8x2 + 2xy + 5x + y2 + y + 6 critical point classification (x, y ) =

Answer 1

Answer:

Therefore the critical point is (x,y)= $(-\frac12,0)$.

Therefore $(-\frac12,0)$ is local minimum.

Step-by-step explanation:

Given function is

$z=f(x,y)=8x^2+2xy+5x+y^2+y+6$

The partial derivatives of z are

$\frac{\partial z}{\partial x}= 8.2x+2y+5=16x+2y+5$

and  

$\frac{\partial z}{\partial y}=2x+2y+1$

To find the critical point, setting the partial derivatives equal to zero.

∴16x+2y+5 =0......(1) and 2x+2y+1=0.......(2)

Now solving the above equation,

Subtract equation (2) from equation (1)

16x+2y+5-( 2x+2y+1)=0

⇒16x+2y+5-2x-2y-1=0

⇒8x+4=0

⇒ 8x = -4

$\Rightarrow x=-\frac{4}{8}$

$\Rightarrow x= -\frac{1}{2}$

Now putting the value of x in (2)

$2.(-\frac12)+2y+1=0$

$\Rightarrow -1 +2y+1=0$

⇒2y=0

⇒y=0

Therefore the critical point is (x,y)= $(-\frac12,0)$.

Second order partial derivatives are

$\frac{\partial^2 z}{\partial x^2 }= 16$ ,$\frac{\partial^2 z}{\partial y^2 }= 2$ and $\frac{\partial^2 z}{\partial x \partial y}= \frac{\partial^2 z}{\partial y \partial x} =2$

The discriminant

$D= \frac{\partial^2 z}{\partial x^2 }.\frac{\partial^2 z}{\partial y^2 }-(\frac{\partial^2 z}{\partial x \partial y})^2$

  $=16+2-(2)^2$

  =14>0

D$(-\frac12,0)$ = 14 >0, Then at $(-\frac12,0)$ is either local minimum or local maximum.

Since  $\frac{\partial^2 z}{\partial x^2 } \right} | _{(-\frac12,0)}= 16>0$, So the function is concave up.

Therefore $(-\frac12,0)$ is local minimum.