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We define $$f(x)=\frac{1}{\sqrt{(x+1/2)^2+y^2+z^2}}$$ and find the maximum and minimum values of $f$ such that $$x^2+y^2+z^2=1$$

Here is what I have so far:

$f(x)$ approaches a maximum as $1/f(x)$ approaches a minimum and vice versa also $1/(f(x))^2$ does the same. So we define a new function $h(x)=1/(f(x))^2=(x+1/2)^2+y^2+z^2$ (side question: is it possible to shift $h(x)$ so it is simply $x^2+y^2+z^2$? or is this invalid?)

So we define the Lagrangian as $$L=((x+1/2)^2+y^2+z^2)+\lambda(x^2+y^2+z^2-1)$$

Then we get the following after taking partial derivatives:

$$L_x=2x+1+2\lambda x=0\\L_y=2y(1+\lambda)=0 \\ L_z=2z(1+\lambda)=0\\L_\lambda=x^2+y^2+z^2-1=0$$

But from here I am stuck, do we want to solve for $x,y,z,\lambda$? And how do we proceed from there if that is the case?

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    $\begingroup$ $$\dfrac{4(x^2+y^2+z^2)+4x+1}4=?$$ $\endgroup$ Mar 24, 2016 at 15:20
  • $\begingroup$ I don't understand what you mean? $\endgroup$
    – Denbo
    Mar 24, 2016 at 15:22
  • $\begingroup$ Expand $(x+1/2)^2$ $\endgroup$ Mar 24, 2016 at 15:25
  • $\begingroup$ $(x+1/2)^2=x^2+x+1/4$ $\endgroup$
    – Denbo
    Mar 24, 2016 at 15:30

1 Answer 1

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You have to solve the four equations simultaneously for $\lambda, x, y, z$. The second equation gives either $\lambda=-1$ or $y=0$. But if $\lambda=-1$ then the first equation becomes $1=0$. Therefore $y=0$, and similarly $z=0$ from the third equation. The fourth equation then gives $x^2=1$, so either $x=1$ or $x=-1$.

You can verify that $x=\pm1$ and $y=z=0$ are the values where $h(x,y,z)$ attains its max and min by using @lab bhattacharjee's hint: Given the constraint $x^2+y^2+z^2=1$, the function $h$ reduces to $x+\frac54$, which attains its max and min under that constraint when $x=\pm1$.

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  • $\begingroup$ Okay I think I am following now just a few comments: 1. Could you instead define $(h(x)=x^2+y^2+z^2)$ and say this has approaches it's max/min as $(x+1/2)^2+y^2+z^2$ does or is that not true? 2. Now we have shown the max and min occurs at $(\pm 1 ,0,0)$ would we just sub that into $f(x,y,z)$ to find max/min values $f(x,y,z)$ takes? $\endgroup$
    – Denbo
    Mar 24, 2016 at 16:35
  • $\begingroup$ You can't use $g(x,y,z) :=x^2+y^2+z^2$ because that will have a different max/min from the original $h(x,y,z):=(x+\frac12)^2+y^2+z^2$. Notice that under the constraint $x^2+y^2+z^2=1$, the function $g$ is constant so its max and min occur at all choices of $x,y,z$ that satisfy the constraint. For your second question, the answer is yes! $\endgroup$
    – grand_chat
    Mar 24, 2016 at 16:41

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