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

MATH105 April 2014

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Question 04 (b)
Let $T(x,y)={\frac {1}{9}}y^{3}+x^{2}-2xy+6x-6y$. Find the maximum and minimum values of $T(x,y)$ on the region $R=\left\{(x,y):{\frac {1}{3}}y^{2}-3\leq x\leq 0\right\}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Sketch the region and check the local max, local min, and points on the boundary.

Hint 2
On the boundary, transform the function to the one with only one variable and then differentiate it to find the minimum and maximum.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\min =-15$.