PLEASE HELP !

Find all critical points of the following functions and determine whether there is a local maximum, a local minimum, or a saddle point at those points.

f(x;y)=(x+y)(xy+1) [Hint: there are two critical points]

Question

PLEASE HELP !

Find all critical points of the following functions and determine whether there is a local maximum, a local minimum, or a saddle point at those points.

f(x;y)=(x+y)(xy+1) [Hint: there are two critical points]

anonymous · Answer

Do you know how to take paritial derivatives?

anonymous · Answer

I would probably multiiply everything out, then take the first and second partial derivatives of the function with respect to x, then take the first and second partial derivatives of the function with respect to y, and then take the partial derivative WRT x then WRT y.

anonymous · Answer

@calmat01 yes what i did was 

Fx= 2xy +1 y'2
Fy= x^2 2yx + 1

but when you get the critical points... how!?

anonymous · Answer

Whoa, a partial with respect to x means the y terms are constant.  So, your 

$$f _{x}(x,y)=2xy+1+y ^{2}$$ and then the partial WRT y is going to be

$$f _{y}(x,y)=x ^{2}+2xy+1$$

anonymous · Answer

So, your partial WRT y is correct.  The partial WRT x needs to be amended.

anonymous · Answer

oops typo

Fx= 2xy +1 + y^2
Fy= x^2 + 2yx + 1

anonymous · Answer

so how do i find the critical numbers?

anonymous · Answer

Now, you set each partial derivative = 0, and solve.  Since we need to find the points that will satisfy both simultaneously, set them equal to each other and the 1 + 2xy terms will cancel out, leaving you with $$x ^{2}=y ^{2}$$

anonymous · Answer

so it's (0,0) ?

anonymous · Answer

@calmat01

anonymous · Answer

You will find that (0,0) will not make either partial derivative 0.

anonymous · Answer

Try something like (-1, 1) or (1,-1)

anonymous · Answer

howw!? :((

anonymous · Answer

We know that x^2=y^2, based on setting our partial derivatives =0, right?

anonymous · Answer

So, we know that $$x ^{2}-y ^{2}=0. $$  But this factors as the difference of two squares.  It factors as (x-y)(x+y)=0, which tells us that either x=y, or x=-y.  Now go back into your partial derivatives and replace x with y or replace x with -y and you will find the values I came up with.  Try it!

anonymous · Answer

@alyannahere  did you get it?

anonymous · Answer

Those are not your only possibilities, but I was just giving you the two that worked to make your partials=0.

anonymous · Answer

righttt. i think i got it? hahahah

anonymous · Answer

thank you!!

anonymous · Answer

Most welcome.