Science:Math Exam Resources/Courses/MATH105/April 2014/Question 04 (b)/Solution 1

We just need to check the local maximum, the local minimum and the points on the boundary. From question $4a$ above, we know that it has only one local minimum $(3,6)$ and no local maximum. Since $(3,6)$ is not on $R$, we only need to find the min and max on the boundary.

Plot of region R

On $C_1$:

$C_1=\{x=0,y\in [-3,3]\},$

$\begin{array}{r}T(x,y)=\frac{1}{9}{y}^3-6y=f(y)f^{\prime }(y)=\frac{1}{3}{y}^2-6\end{array}$

The critical points from above occur when $\frac{1}{3}{y}^2-6=0$ which is at $y=\pm \sqrt{18}=\pm 3\sqrt{2}$, neither of which are in $[-3,3]$. Thus, we just test the endpoints: $f(-3)=15$ and $f(3)=-15$. These are two values we will consider later.

On $C_2$:

$C_2=\{x=\frac{1}{3}{y}^2-3,y\in [-3,3]\},$

$T(x,y)=\frac{1}{9}{y}^3+{(\frac{1}{3}{y}^2-3)}^2-2(\frac{1}{3}{y}^2-3)y+6(\frac{1}{3}{y}^2-3)-6y=\frac{y^4}{9}-\frac{5}{9}{y}^3-9=g(y)$

$g^{\prime }(y)=\frac{4}{9}{y}^3-\frac{15}{9}{y}^2=\frac{y^2}{9}(4y-15)=0$ at $y=0$ and $y=15/4$. As $15/4$ is not in $[-3,3]$ we ignore it. We now compute $g(0)=-9$ and the value at the endpoints with $g(-3)=15$ and $g(3)=-15$.

From all of this, we find $\max =15$ and $\min =-15$.