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

MATH105 April 2015

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Question 03 (a)
Let $f(x,y)=(x-1)^{2}+(y+1)^{2}$. 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.

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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.

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