Science:Math Exam Resources/Courses/MATH105/April 2015/Question 03 (a)/Solution 1

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 $f(x,y)=(x-1)^{2}+(y+1)^{2}$ and the constraint is $g(x,y)=x^{2}+y^{2}=4$. This gives

${\begin{cases}(2(x-1),2(y+1))=\lambda (2x,2y)\\g(x,y)=4.\end{cases}}$

The first equation produces the system

${\begin{cases}2x-2&=\lambda \cdot 2x\\2y-2&=\lambda \cdot 2y,\end{cases}}$

and isolating $\lambda$ we find that

${\begin{aligned}\lambda ={\frac {2x-2}{2x}}&={\frac {2y+2}{2y}}\\1-{\frac {1}{x}}&=1+{\frac {1}{y}}\\x&=-y.\end{aligned}}$

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

$f({\sqrt {2}},-{\sqrt {2}})=({\sqrt {2}}-1)^{2}+(-{\sqrt {2}}+1)^{2}=2-2{\sqrt {2}}+1+2-2{\sqrt {2}}+1=6-4{\sqrt {2}},$

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,

${\begin{aligned}\max _{x^{2}+y^{2}=4}f(x,y)&={\color {blue}6+4{\sqrt {2}}}\\\min _{x^{2}+y^{2}=4}f(x,y)&={\color {blue}6-4{\sqrt {2}}}.\end{aligned}}$