Science:Math Exam Resources/Courses/MATH105/April 2013/Question 05 (b)

MATH105 April 2013

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[hide] Question 05 (b)
Use the method of Lagrange multipliers to find the maximum and minimum values of $\displaystyle f(x,y)=6y-y^{3}-3x^{2}y$ on the circle $\displaystyle x^{2}+y^{2}=4$ 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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[show] Hint
Let $\displaystyle g(x,y)=x^{2}+y^{2}-4$ be the constraint function. We thus need to solve the system of equations given by ${\begin{aligned}g(x,y)&=0\\\nabla f(x,y)&=\lambda \nabla g(x,y)\end{aligned}}$

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[show] Solution
Let $\displaystyle g(x,y)=x^{2}+y^{2}-4$ be the constraint function. We thus need to solve the system of equations given by ${\begin{aligned}g(x,y)&=0\\\nabla f(x,y)&=\lambda \nabla g(x,y)\end{aligned}}$ Plugging in our values gives ${\begin{aligned}x^{2}+y^{2}-4&=0\\(-6xy,6-3y^{2}-3x^{2})&=\lambda (2x,2y)\end{aligned}}$ Expanding the second equation gives the following three equations ${\begin{aligned}x^{2}&=4-y^{2}\\-6xy&=2\lambda x\\6-3y^{2}-3x^{2}&=2\lambda y\end{aligned}}$ Substituting the first equation in the third equation gives $\displaystyle 6-3y^{2}-3(4-y^{2})=2\lambda y$ which simplifies to $\displaystyle -6-3y^{2}+3y^{2}=2\lambda y$ Reducing gives $\displaystyle -3=\lambda y$ . This tells us that $\displaystyle \lambda \neq 0$ and thus we can divide by it to get that ${\frac {-3}{\lambda }}=y$ . Looking at the second equation from the group of three above gives us that $\displaystyle 0=2\lambda x+6xy=x(2\lambda +6y)$ Case 1: x = 0 Plugging x = 0 into $\displaystyle g(x,y)$ gives $\displaystyle y=\pm 2$ . Case 2: x ≠ 0 In this case we divide the equation above by x and obtain $2\lambda =-6y=-6\left({\frac {-3}{\lambda }}\right)={\frac {18}{\lambda }}$ Solving gives $\lambda =\pm 3$ . Thus $y={\frac {-3}{\lambda }}=\mp 1$ . Plugging these y values into $\displaystyle g(x,y)$ gives $x=\pm {\sqrt {3}}$ . Compare function values at all candidate points We found the following six solutions: $\displaystyle (0,-2),(0,2),(-{\sqrt {3}},-1),({\sqrt {3}},-1),(-{\sqrt {3}},1),({\sqrt {3}},1)$ Plugging in each of the values gives ${\begin{aligned}f(0,2)&=6(2)-(2)^{3}-3(0)^{2}(2)=4\\f(0,-2)&=6(-2)-(-2)^{3}-3(0)^{2}(-2)=-4\\f(-{\sqrt {3}},-1)&=6(-1)-(-1)^{3}-3(-{\sqrt {3}})^{2}(-1)=4\\f({\sqrt {3}},-1)&=6(-1)-(-1)^{3}-3({\sqrt {3}})^{2}(-1)=4\\f(-{\sqrt {3}},1)&=6(1)-(1)^{3}-3(-{\sqrt {3}})^{2}(1)=-4\\f({\sqrt {3}},1)&=6(1)-(1)^{3}-3({\sqrt {3}})^{2}(1)=-4\end{aligned}}$ and thus the minimum value is -4 and the maximum value is 4.