MATH105 April 2015
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Question 03 (a)

Let $f(x,y)=(x1)^{2}+(y+1)^{2}$.
Use the method of Lagrange multipliers to find the maximum and minimum values of $f(x,y)$ on the circle $x^{2}+y^{2}=4$. A solution that does not use the method of Lagrange multipliers will receive no credit, even if the answer is correct.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Recall that the method of Lagrange multipliers involves solving the following system of equations:
 ${\begin{cases}(f_{x}(x,y),f_{y}(x,y))=\lambda (g_{x}(x,y),g_{y}(x,y))\\g(x,y)=c,\end{cases}}$
where $f(x,y)$ is the function to maximize/minimize, subject to the constraint $g(x,y)=c$, and $\lambda$ is the Lagrange multiplier.
Solve this system for $x$, $y$, and $\lambda$, and plug $x,y$ into $f$ to find the maximum/minimum values.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
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Solution

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We apply the method of Lagrange multipliers,
 ${\begin{cases}(f_{x}(x,y),f_{y}(x,y))=\lambda (g_{x}(x,y),g_{y}(x,y))\\g(x,y)=c,\end{cases}}$
where $f(x,y)=(x1)^{2}+(y+1)^{2}$ and the constraint is $g(x,y)=x^{2}+y^{2}=4$. This gives
 ${\begin{cases}(2(x1),2(y+1))=\lambda (2x,2y)\\g(x,y)=4.\end{cases}}$
The first equation produces the system
 ${\begin{cases}2x2&=\lambda \cdot 2x\\2y2&=\lambda \cdot 2y,\end{cases}}$
and isolating $\lambda$ we find that
 ${\begin{aligned}\lambda ={\frac {2x2}{2x}}&={\frac {2y+2}{2y}}\\1{\frac {1}{x}}&=1+{\frac {1}{y}}\\x&=y.\end{aligned}}$
Substituting $x=y$ into the equation of the constraint $g(x,y)=4,$ we obtain
 ${\begin{aligned}x^{2}+y^{2}=x^{2}+(x)^{2}=2x^{2}&=4\\x&=\pm {\sqrt {2}}\\y&=x=\mp {\sqrt {2}}.\end{aligned}}$
Finally, comparing the values of $f$ at these points, we see that
 $f({\sqrt {2}},{\sqrt {2}})=({\sqrt {2}}1)^{2}+({\sqrt {2}}+1)^{2}=2+2{\sqrt {2}}+1+2+2{\sqrt {2}}+1=6+4{\sqrt {2}}$
and
 $f({\sqrt {2}},{\sqrt {2}})=({\sqrt {2}}1)^{2}+({\sqrt {2}}+1)^{2}=22{\sqrt {2}}+1+22{\sqrt {2}}+1=64{\sqrt {2}},$
from which we conclude that the maximum occurs when $x={\sqrt {2}}$ and $y=+{\sqrt {2}}$, while the minimum occurs when $x=+{\sqrt {2}}$ and $y={\sqrt {2}}$. In summary,
 ${\begin{aligned}\max _{x^{2}+y^{2}=4}f(x,y)&={\color {blue}6+4{\sqrt {2}}}\\\min _{x^{2}+y^{2}=4}f(x,y)&={\color {blue}64{\sqrt {2}}}.\end{aligned}}$
