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{\bf Exam Question

Topic: CriticalPoints}

Find and classify the critical points of the function
$$f(x,y)=3x^4+12xy+4y^3.$$ Calculate the value of the function at
each of the critical points. \vspace{0.5in}

{\bf Solution}

$$f(x,y)=3x^4+12xy+4y^3;\ f_x=12x^3+12y;\ f_y=12x+12y^2$$ So the
partial derivatives are zero when $x^3+y=0;\ x+y^3=0.$

Substituting for $x$ gives $-y^6+y=0$ i.e. $y(1-y^5)=0.$

The only real solutions are $y=0,\ y=1.$

When $y=0,\ x=0$ and when $y=1,\ x=-1.$

Calculating the second partial derivatives gives

$$f_{xx}=36x^2;\ f_{yy}=24y;\ f_{xy}=12.\ \mathrm{So}\
\Delta=f_{xy}^2-f_{xx}f_{yy}=144-864x^2y $$

So at $(0,0), \Delta=144>0$ so $(0,0)$ is a saddle point.

At $(-1,1), \Delta=144-864<0$ so $(-1,1)$ is a local minimum
$(f_{xx}>0).$

The values of $f$ at the critical points are $f(0,0)=0;\
f(-1,1)=-5.$


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