Pretty easy relative max problem calc 3

[Mdhiggenz]

Homework Statement

Locate all relative max,min,and saddle points if any.

f(x,y)=x²+xy+y²-3x

fx=2x+y-3

fy=x+2y

Skipping some algebra I get the critical points (2,-1)

fxx=2
fyy=2

d=fxx*fyy-f(x0,y0)²

d=4-9=-5

I know I'm messing up at f(x0,y0)

I'm simply plugging in my c.p points (2,-1) into

f(x,y)=x²+xy+y²-3x

but seem to keep getting -3

While the value 1 is suppose to come out

Homework Equations

The Attempt at a Solution

 

[Dick]

Homework Statement

Locate all relative max,min,and saddle points if any.

f(x,y)=x²+xy+y²-3x

fx=2x+y-3

fy=x+2y

Skipping some algebra I get the critical points (2,-1)

fxx=2
fyy=2

d=fxx*fyy-f(x0,y0)²

d=4-9=-5

I know I'm messing up at f(x0,y0)

I'm simply plugging in my c.p points (2,-1) into

f(x,y)=x²+xy+y²-3x

but seem to keep getting -3

While the value 1 is suppose to come out

Homework Equations

The Attempt at a Solution

It looks like you have your formula for d wrong. It's supposed to be d=fxx*fyy-(fxy)^2. fxy is the mixed second derivative.

 

[Mdhiggenz]

I'm a bit confused what do you mean mixed second derivative?

 

[Dick]

I'm a bit confused what do you mean mixed second derivative?

First take the derivative of f with respect to x to get fx. Then take the derivative of fx with respect of y to get fxy. That's the mixed second derivative. Or you can do it in the other order and get fyx. They should be the same.

 

[Mdhiggenz]

Got it, thanks Dick.