Finding minimum value using Lagrange multipliers

Question

I need to find the minimum of $f(x,y,z) = x^2 + y^2 + z^2$ subject to the constraints $(x-2)^2 + y^2 + z^2 = 1$ and $x + z = 3$.

I've got the following equations set up (using Lagrange multipliers):

$$2x = 2(x-2)\lambda+\mu$$ $$2y = 2y\lambda$$ $$2z = 2z\lambda+\mu$$

I suspect the answer is $x=2, y=0, z=1$, but I don't know how to arrive at it.

Is my problem set up correctly? If so how do I solve for $x, y, z, \lambda, \mu$?

Answer 1

From the second equation either $y=0$ or $\lambda=1$. The latter results in an inconsistent system, so set $y=0$ and continue from there. The second constraint allows you to eliminate either $x$ or $z$, and after eliminating $y$ the second constraint becomes a simple quadratic equation in the remaining variable. Since the objective function is the squared distance from the origin, it should be easy to pick out which of the two solutions is minimal.

Of course, like many of the exercises given for practice in the Lagrange multiplier method, this problem can be solved without using the method. Geometrically speaking, you’re being asked to find the nearest point to the origin of the intersection between a sphere and plane. This intersection is a circle. Parameterizing this circle reduces the problem to a straightforward single-variable minimization.

Answer 2

The reason you are having trouble setting up the LaGrange multiplier method on this problem is, because the fact that $2y=2y\lambda$ does not imply that $\lambda=1$. The correct conclusions is that, $2y(1-\lambda)=0$ so either $\lambda=1$ or $y=0$. Since it is the latter that happens, you cannot obtain the value of $\lambda$ from it.

In fact, there is a way of solving this problem without LaGrange multipliers. So feel free to use any part of this solution that you feel would simplify your approach.

The fact that $(x-2)^2+y^2+z^2=1$ means you can add and subtract that equation from your $f(x)$ to see that in fact, $$ f(x)=x^2+y^2+z^2-((x-2)^2+y^2+z^2)+1=4x-4+1=4x-3 $$

So now, you're looking to minimize $4x-3$ with respect to the conditions above, which is equivalent to minimizing just $x$.

Finally, since $x+z=3$, then $z=3-x$, so it's equivalent to maximizing $z$. (We could also continue by minimizing $x$, but the arithmetic would become more complicated.) Replacing $x=3-z$ in the first constraint equation, you will see that it becomes $$1=(z-1)^2+y^2+z^2=2z^2-2z+y^2+1$$ $$0=2z(z-1)+y^2$$

Since $y^2$ is positive when $y\neq 0$, then the only way this sum can equal zero is when $y=0$.

At this point, it's pretty clear that this equation has only two roots, $z=0$ and $z=1$, and those are the only two that make this equation true. Out of these, $z=1$ is the bigger solution, and after back-substituting we see that your proposed answer is correct.