Science:Math Exam Resources/Courses/MATH105/April 2014/Question 03 (a)

MATH105 April 2014

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[hide] Question 03 (a)
Use the method of Lagrange multipliers to find the maximum and minimum values of $(x+1)^{2}+(y-2)^{2}$ on the circle $x^{2}+y^{2}=125$ . 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
The method of Lagrange multipliers states that the extreme values of the objective function $f(x,y)$ subject to the constraint $g(x,y)=0$ will occur at the solutions to the system: $\nabla f=\langle f_{x},f_{y}\rangle =\lambda \langle g_{x},g_{y}\rangle =\lambda \nabla g$ .

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[show] Solution
The object function is $f(x,y)=(x+1)^{2}+(y-2)^{2}$ and the constraint is $x^{2}+y^{2}=125$ , i. e. $g(x,y)=x^{2}+y^{2}-125=0$ . By the method Lagrange multipliers, set $\nabla f=\lambda \nabla g,g=0$ which tells us $\langle 2(x+1),2(y-2)\rangle =\lambda \langle 2x,2y\rangle ,\quad x^{2}+y^{2}-125=0$ . Looking at the vector equation in components, we have that ${\begin{aligned}2(x+1)=2\lambda x&\implies x+1=\lambda x\quad (1)\\2(y-1)=2\lambda y&\implies y-2=\lambda y.\quad (2)\end{aligned}}$ Provided we don't divide by zero (so that $y\neq 2$ ), we can divide (1) by (2) to yield ${\frac {x+1}{y-2}}={\frac {\lambda x}{\lambda y}}={\frac {x}{y}}\implies xy+y=xy-2x\implies y=-2x$ where the first implication came by cross multiplying. If $y=x$ then from the constraint $x^{2}+y^{2}-125=0$ we must have $x^{2}+(-2x)^{2}-125=5x^{2}-125=0\implies x=\pm 5$ and $y=-2x=\mp 10$ . Evaluating $f(5,-10)=(5+1)^{2}+(-10-2)^{2}=180$ and $f(-5,10)=(-5+1)^{2}+(10-2)^{2}=80$ . We still need to consider the possibility that $y=2$ . If $y=2$ then (2) reads $0=2\lambda \implies \lambda =0$ . If $\lambda =0$ then (1) tells us that $x+1=0$ so that $x=-1$ . However, $(-1,2)$ does not satisfy $g(x,y)=0$ so this is not a valid solution to the Lagrange system. Overall have found that the maximum value is 180 and the minimum value is 80.