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Math Help - critical points

  1. #1
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    critical points

    I found two different approaches for solving this one. Still confused. What do you suggest to look for the first and then second derivative. Or anything else is on your mind. Please take a look. Thx a lot !


    For each of the following functions, find the critical points and classify them as local
    maximum, local minimum, saddle point, or "can't tell":
    a)
    xy^(2)+ x^(3)y xy b) x^(2) 6xy + 2y^(2) +10x + 2y 5
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  2. #2
    Senior Member
    Joined
    Dec 2008
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    For both, we may notice that the domain is the entire xy-plane and contains no boundary points. We are therefore left with points at which \nabla z=0. For (a), this amounts to

    \begin{aligned}<br />
\nabla (xy^2+x^3y-xy)&=0\\<br />
\frac{\partial}{\partial x}(xy^2+x^3y-xy)\textbf i\,+\frac{\partial}{\partial y}(xy^2+x^3y-xy)\textbf j\,&=0\\<br />
(y^2+3x^2y-y)\textbf i\,+(2xy+x^3-x)\textbf j\,&=0.<br />
\end{aligned}<br />

    When y=0 at one of these critical points,

    \begin{aligned}<br />
x^3-x&=0\\<br />
x(x^2-1)&=0\\<br />
x(x+1)(x-1)&=0,<br />
\end{aligned}

    and we must have x=-1,\,0,\mbox{ or }1. Similarly, when x=0, we must have

    \begin{aligned}<br />
y^2-y&=0\\<br />
y(y-1)&=0,<br />
\end{aligned}

    and therefore y=0\mbox{ or }1. When neither x nor y equals 0, we may divide by x and y in both:

    \begin{aligned}<br />
y+3x^2-1&=0\\<br />
2y+x^2-1&=0.<br />
\end{aligned}<br />

    Multiplying both sides of the top equation by 2 and subtracting the bottom gives

    5x^2-1=0,

    and therefore x=\pm\frac{1}{\sqrt{5}},\,y=\frac{2}{\sqrt{5}}. Substitution confirms that all of these are actual solutions of \nabla z=0. To determine the nature of each critical point, remember that the discriminant D is defined as

    \begin{aligned}<br />
D&=\frac{\partial^2z}{\partial x^2}\cdot\frac{\partial^2z}{\partial y^2}-\left(\frac{\partial^2z}{\partial x\partial y}\right)^2\\<br />
&=f_{xx}f_{yy}-f_{xy}^2,<br />
\end{aligned}<br />

    and that

    1. D>0 confirms a local extremum and
    2. D<0 confirms a saddle point, while
    3. D=0 tells us nothing.

    Some textbooks define the discriminant as D=f_{xy}^2-f_{xx}f_{yy}. When this happens, then D<0 confirms extrema and D>0 confirms saddle points. I'm not sure which sign your teacher uses.
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