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 $\mathrm{\nabla }f=\lambda \mathrm{\nabla }g,g=0$ which tells us $⟨2(x+1),2(y-2)⟩=\lambda ⟨2x,2y⟩,x^2+y^2-125=0$.

Looking at the vector equation in components, we have that

$\begin{array}{rl}2(x+1)=2\lambda x& ⟹x+1=\lambda x(1)\\ 2(y-1)=2\lambda y& ⟹y-2=\lambda y.(2)\end{array}$

Provided we don't divide by zero (so that $y\ne 2$), we can divide (1) by (2) to yield $\frac{x+1}{y-2}=\frac{\lambda x}{\lambda y}=\frac{x}{y}⟹xy+y=xy-2x⟹y=-2x$ where the first implication came by cross multiplying.

If $y=x$ then from the constraint $x^2+y^2-125=0$ we must have $x^2+(-2x{)}^2-125=5x^2-125=0⟹x=\pm 5$ 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 $0=2\lambda ⟹\lambda =0$. 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.