Critical points of 2-Variable Function

Saladsamurai
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Homework Statement


Find local max/min, and saddle points (if any) of

f(x,y)=x^2+y^2+x^2y+4

This should be simple, but I am having algebra-block on solving the partial derivatives to find the critical points.

f_x=2x+2xy=0 (1)
f_y=2y+x^2=0 (2)

If I multiply the second equations by -x an add it to the second and solve for x, I get

x={0,+sqrt2, -sqrt2}

But for some reason I cannot figure out how to solve equation 2 ?

Why am I retarded?
 
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Well, for x=0 equation 2 gives: 2y+(0)^2=0 \Rightarrow y=0 How about for x=+sqrt2?
 
I guess I don't understand this.

I want to know at what points the slope of the tangent lines is 0.

Now, I need values of x and y that satisfy both equations simultaneously. I am used to solving the equations simultaneously, not by assigning specific values to x or y.

I am not sure why that bothers me so much. But as you say:

if x=0 then eq 2 is satisfied by y=0, thus (0,0) is critical

if x=+ or - sqrt2 eq 2 is satisfied by y=-1, thus (+sqrt2, -1) and (-sqrt2, -1) are critical, yes?
 
Yes, you can double check that those points satisfy both equations by substituting each point into them.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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