Find and analyze the local extrema of the function f(x,y)=x 2 +xy+y 2 now I know

Question

Find and analyze the local extrema of the function f(x,y)=x2+xy+y2

now I know how to find critial points and found critical points to be (0,0)

Solution:

fx(x, y) = 2x + y and fy(x, y) = x + 2y. Setting these equations equal to 0, we have two
equations:
2x + y = 0 and x + 2y = 0
We can solve for x and y by multiplying the second equation by -2 and adding the
equations. Doing this, we have that -3y = 0, and so y = 0. Plugging that into the first
equation, we see that x = 0. Thus, the only critical point is (0, 0).

Notice that f(0, 0) = 0. Thus, if f(x, y) is always positive or zero near (0, 0), then we
would have that (0, 0) is a local minimum. If f(x, y) is always negative or zero near (0, 0),
then we would have a local maximum.

My question:

1) "if f is always positive or zero near (0,0), then (0,0) is a local minimum; if f is always negative or near zero near (0,0), it is a local maximum". in this problem, we found that f(0,0) resulted in 0 (ie: (0)2+(0*0)+(0)2 = 0) now how do we know if its local minimum or maximum by this? i mean "if f is zero near (0,0)" applies to both situations... and in this problem, f is not positive or negative to be certain, I do not know if its local maximum or minimum.

2) what does critical point have anything to do with this problem though? all that matters is if f(x,y) = x2+xy+y2 yields positive or negative if we input (0,0) for x and y no?

3) the solution says f(x,y) = x2+xy+y2 = (x+(1/2)y)2 + (3/4)y2 by completing the square. Please provide me a (clear) step by step without missing any steps to this...

Explanation / Answer

Example of multiple points:

Let f(x,y) = 10 + x^3 + y^3 - 3xy

fx(x,y) = 3x^2 - 3y = 0 ....(1)
fy(x,y) = 3y^2 -3x = 0 ....(2)
fxx(x,y) = 6x
fyy(x,y) = 6y
fxy (x,y) = -3

critical points:
3x^2 - 3y = 0 OR x^2 = y
3y^2 - 3x = 0 OR y^2 = x

Putting y = (x^2) in the second equation

(x^2)^2 = x
x^4 = x
x^3(x-1) = 0
x = 0 or x = 1

when x = 0, y = x^2 = 0
when x = 1, y = x^2 = 1

Thus points are (0,0) abd (1,1)

D(x,y) = fxx(x,y)*fyy(x,y) - fxy(x,y) ^ 2 = 6x6y - (-3)^2
=36xy - 9

at (0,0), D = -9 <0 i.e (0,0) is a saddle point

at (1,1), D = 36-9 = 27 > 0 and fxx(1,1) = 6 >0
Thus (1,1) is a local minima

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.