- #1

giacomh

- 36

- 0

**So, I technically got this problem correct (our professor gave us the answers but not the total solution). My question is more of a logic one- an answer that I'll need to apply this concept to problems on the exam. I included it at the bottom of the post, after I explained the problem**

Find local max and min of f(x,y)=xy

^{2}-yx

^{2}+xy

## Homework Equations

fx=y

^{2}-2yx+y fy=2xy-x

^{2}+x

fxx=-2y fxy=2y-2x+1 fyy=2x

## The Attempt at a Solution

y

^{2}-2yx+y=0

y(y-2x+1)=0

y=0 OR y-2x+1=0

2xy-x

^{2}+x

x(2y-x+1)=0

x=0 OR 2y-x+1=0

Plug y=0 into 2y-x+1=0

x=1

Plug x=0 into y-2x+1=0

y=-1

Therefore, -y=x. Plug this into y-2x+1=0

y-2(-y)+1=0

y+2y+1=0

y=-1/3 x=1/3

**Critical Points:**

(1,0)

(0,1)

(1/3,-1/3)

**(I know how to solve for min and max from here, I don't need to type it out)

**My question:**

I don't understand the relationship between the fy and fx derivatives. For instance, I understand why x=0 OR 2y-x+1=0 (from fx), but I don't understand why you would assume that plugging y=0 into 2y-x+1=0 is correct? Why does 2y-x+1=0 when y=0? Rather, why does plugging y=0 into 2y-x+1=0 give you a point on the graph, when 2y-x+1 isn't the derivative itself, only only part of the factored equation? Why don't you take into account the y factored out of (2y-x+1), since you can't factor out a zero (because you're assuming y=0). If x=0 and y=0 are both solutions, why isn't (0,0) a critical point?

Furthermore, I understand the algebra behind plugging -y=x into y-2x+1=0 to get -1/3 and 1/3, but I would never think to do that. How do you know there is a third answer, when you already found 2 answers from the factored y-2x+1=0 (or 2y-x+1=0).

Its just hard for me to visualize what's going on!