Science:Math Exam Resources/Courses/MATH105/April 2014/Question 03 (a)/Solution 1

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.