Solving Lagrange Multipliers: Find Extrema of Distance from (1,2,3) to Sphere

[ctb94]

Hello,

I am having a bit of trouble with the Lagrange multiplier method. My question is:

Use the Lagrange multiplier method to find the extrema points of the distance from the point (1,2,3) to the surface of the sphere {x}^{2}+{y}^{2}+{z}^{2}=4. Find the possible values for of \lambda.

This is what I have so far:

\nablad=\lambdaq

d(x,y,z)={(x-1)}^{2}+{(y-2)}^{2}+{(z-3)}^{2}

2(x-1)=\lambda2x
2(y-2)=\lambda2y
2(z-3)=\lambda2z

This is where I become stuck on the problem. Thank you in advance for any help with this problem!

 

[MarkFL]

I would begin by observing that the given sphere of radius 2 is centered at the origin, and the point $(1,2,3)$ is $\sqrt{14}$ units from the origin, so our extrema will have values of $\sqrt{14}\pm2$.

Now, we are given the objective function (let's use the square of the distance for simplcity):

\(\displaystyle f(x,y,z)=(x-1)^2+(y-2)^2+(z-3)^2\)

subject to the constraint:

\(\displaystyle g(x,y,z)=x^2+y^2+z^2-4=0\)

So, Lagrange gives us the system (after dividing through by 2 in each case):

\(\displaystyle x-1=\lambda x\)

\(\displaystyle y-2=\lambda y\)

\(\displaystyle z-3=\lambda z\)

Now, the first two equations imply:

\(\displaystyle \frac{x-1}{x}=\frac{y-2}{y}\implies y=2x\)

And the first and third equations imply:

\(\displaystyle \frac{x-1}{x}=\frac{z-3}{z}\implies z=3x\)

Substituting these values into the constraint, what do you find?

 

[ctb94]

When substituting these values back into the constraint, I got x=+/-sqrt(2/7). Is this correct? If so, do I then substitute this back into y=2x and z=3x?

 

[MarkFL]

When substituting these values back into the constraint, I got x=+/-sqrt(2/7). Is this correct? If so, do I then substitute this back into y=2x and z=3x?

Yes, you are correct on both counts...how many critical points will this generate?

 

[ctb94]

I believe it should be six critical points then, correct?

 

[MarkFL]

I believe it should be six critical points then, correct?

There are 2 possible values for each of $x$, $y$ and $z$, so the fundamental counting principle says there will be 8 permutations, in much the same way that a 3 digit binary number can have one of 8 possible values.

So, what you want to do is test all of these 8 critical points in your objective function, and then choose only the 2 points that result in your extrema, that is, the point that results in the smallest value and the point that results in the largest value.

 

[ctb94]

That makes complete sense! Thank you so much for making this concept so much easier to understand!

 

[MarkFL]

Looking at the form of the objective function, we should easily be able to see that the maximum occurs when all the variables are negative, and the minimum occurs when they are all positive...let's see:

\(\displaystyle f_{\min}\left(\sqrt{\frac{2}{7}},2\sqrt{\frac{2}{7}},3\sqrt{\frac{2}{7}}\right)=14\left(\sqrt{\frac{2}{7}}-1\right)^2=14\left(\frac{2}{7}-2\sqrt{\frac{2}{7}}+1\right)=18-4\sqrt{14}=\left(\sqrt{14}-2\right)^2\)

\(\displaystyle f_{\max}\left(-\sqrt{\frac{2}{7}},-2\sqrt{\frac{2}{7}},-3\sqrt{\frac{2}{7}}\right)=14\left(\sqrt{\frac{2}{7}}+1\right)^2=14\left(\frac{2}{7}+2\sqrt{\frac{2}{7}}+1\right)=18+4\sqrt{14}=\left(\sqrt{14}+2\right)^2\)

These values do in fact agree with the values we predicted at the beginning (noting that our objective function is the square of the distance).