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The function are given below $$f(x,y)=xy^2-x^2-x$$

I have done the below steps $$f_x(x,y)=y^{2}-2x-1$$ $$f_y(x,y)=2xy$$ Therefore the critical points are $$(-0.5,0)(0,1)(0,-1)$$ Sub the contraints into the original function $$H(x)=-x^3-x^2+x$$ $$\text{Derivative of }H(x)=-3x^2+2x+1$$ Let $\text{Derivative of }H(x) =0$, $x=-1$ and $1/3$

How do i proceed further from here?

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I can't understand what you're doing after finding the critical points. You're supposed to find the hessian matrix at each of these points to check what happens. – Git Gud Apr 2 '14 at 11:16
there is a constraint and i have to take it into account – Jake Mitch Apr 2 '14 at 11:18
@user136934 But the critical points satisfy the constrain, so you can proceed as if there was no constraint. – Git Gud Apr 2 '14 at 11:20
so i need to do the steps below? 1)use the values of x in the derivative of H(x) and sub into the actual H(x) function 2)use the critical points and sub into f(x,y) 3) compare with (1) and (2) to see which is the abs max and abs min? – Jake Mitch Apr 2 '14 at 11:23
It seems to me you're using something that I don't know, that's why I can't understand what did after finding the critical points. I can't help you. But what I said is still true, you can just find the hessian matrix. – Git Gud Apr 2 '14 at 11:29

$xy^2-x^2-x=x\cdot(y^2-x-1)$, so we can make estimates based on whether $x>0$ or $x<0$:

$$\begin{align}x\cdot(y^2-x-1)&\ge\min(\sqrt2\cdot(0-\sqrt2-1),\,-\sqrt2\cdot(2+\sqrt2-1))=-2-\sqrt2\\\\x\cdot(y^2-x-1)&\le\max(x\cdot((2-x^2)-x-1),\,x\cdot(0-x-1))\\&=\max(-x^3-x^2+x,\,-x^2-x)\end{align}$$ The minimum estimate is clear, for the maximum take the derivatives: $$\begin{align}(-3x^2-x^2+x)'&=-3x^2-2x+1=(1-3x)(1+x)\\(-x^2-x)'&=-2x-1\end{align}$$ The first is $0$ when $x=\frac13$ or $x=-1$, the second is $0$ when $x=-\frac12$. Checking the local extremes and ends of the interval $[-\sqrt2,\sqrt2]$ you should get that the maximum of the second estimate higher than the maximum of the first and it's only for $x=-\frac12$, so $$x\cdot(y^2-x-1)\le-(-\!\tfrac12)^2-(-\!\tfrac12)=\tfrac14$$

So the minimum is $-2-\sqrt2$, because it's attained for $(x,y)=(\sqrt2,0)$ and the maximum is $\frac14$, because it's attained for $(x,y)=(-\!\frac12,0)$.

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my abs min coincides with yours but i can't figure out how you get the abs max. when i sub (-0.5,0) into f(x,y) i received -0.75 – Jake Mitch Apr 2 '14 at 12:37
also i don't understand how your method work as following the absolute extrema on closed bounded region, i understand that i must find the critical points in the interior which is f(x,y) and extreme values in the boundary which is x^2+y^2<=2 – Jake Mitch Apr 2 '14 at 12:40
@JakeMitch I'm not using any advanced theory. Just simple estimates. If $x^2+y^2\le2$, then $-\sqrt2\le x,y\le\sqrt2$. So for $x>0$ we can minimize $x\cdot(y^2-x-1)$ by setting $x$ before the dot to be the greatest possible and $y^2-x-1$ after the dot to be least possible. Similarly the other estimate if $x<0$, then $x$ is least when $x=-\sqrt2$ and $y^2-x-1$ is greatest when $x=-\sqrt2$ and $y=\sqrt2$. – user2345215 Apr 2 '14 at 12:51

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