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 $C_{1}$:

$C_{1}=\{x=0,y\in [-3,3]\},$

${\begin{aligned}T(x,y)={\frac {1}{9}}y^{3}-6y=f(y)\quad \quad f'(y)={\frac {1}{3}}y^{2}-6\end{aligned}}$

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 $C_{2}$:

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

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

$g'(y)={\frac {4}{9}}y^{3}-{\frac {15}{9}}y^{2}={\frac {y^{2}}{9}}(4y-15)=0$ 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 $\min =-15$.