Lagrange Multipliers Method of solving Question

Question

Find the maximum and minimum values of $f(x, y) = x^2 + y^2$

subject to the constraint $x^2 − 2x + y^2 − 4y = 0$

So I have to use lagrange multipliers

$ \nabla f(x,y) = \lambda\nabla g(x,y) $

$$ \nabla f(x,y)= <2x,2y> $$

$$ \lambda\nabla g(x,y) = \lambda<2x-2,2y-4> $$

making x and y the subjects gives:

$$ x = x\lambda - \lambda$$ $$ y = y\lambda - 2\lambda$$

I need to plug these x and y's back into the original constraint ($x^2 − 2x + y^2 − 4y = 0$), but I'm not sure how to go about it without having a very big equation. Any help here would be much appreciated.

Answer 1

You can divide one equation by the other. For this purpose we assume, for the moment, that the denominators are unequal $0$.

$\frac{x}{y}=\frac{x-1}{y-2}$

Multipliying both sides by $y$ and $y-2$

$x(y-2)=y(x-1)$

$xy-2x=yx-y$

Substracting xy on both sides.

$-2x=-y$

$2x=y$

Plugging in the term for y into the constraint.

$x^2-2x+(2x)^2-4\cdot 2x=0$

$5x^2=10x$

$x_1=2$ and $x_2=0$

Thus $y_1=4$ and $y_2=0$

This are the two critical points. Now you can use the bordered Hessian to determine if one is a maximum or a minimum.

---

The bordered Hessian is

H = $\left( \begin{array}{} 0 & g_{x} & g_{y} \\ g_{x} & L_{xx} & L_{yy} \\ g_{y} & L_{yx} & L_{yy}\end{array}\right)$

with $g=x^2-2x+y^2-4y$

And $L=x^2+y^2+\lambda(x^2-2x+y^2-4y)$

For example $L_{xx}=2+2\lambda$. You differentiate $L$ twice w.r.t. $x$. Thus you have to calculte $\lambda_1$ and $\lambda_2$ for the two different critical points. When you have calculated all values of H then you calculate the determinant of H.

If $det \ H(x_0,y_0) >0 \Rightarrow \texttt{it´s a (local) maximum}$

If $det \ H(x_0,y_0) <0 \Rightarrow \texttt{it´s a (local) minimum}$

Answer 2

The constraint defines the circle $$(x-1)^2+(y-2)^2=5$$ going through the origin, and you are told to find the two points on this circle which are at maximal, resp., minimal distance from the origin. The nearest point is $(0,0)$, of course, and the farthest point is the other end of the diameter through $(0,0)$, i.e., the point $(2,4)$. It follows that $\max f$ under the given constraint is $20$, and $\min f$ is $0$.

Answer 3

For this problem, you may not use Lagrange multiplier. The objective function is a cone and the constraint is, as you said, circle with center (1,2) and radius $\sqrt5$. So the max-min is on the curve which is the intersection between the cone and the cylinder. You can vision that there are two points on the curve corresponding to min (close to the origin) and max along the line (0,0) to (1,2), or $y=2x$

Plugging the above line into the constraint, you can get x=0 and x=2. So the minimum is at x=0 with value 0 and the maximum is at 2 with the value 20.