# Homework Help: Min and max value of r

1. Apr 27, 2016

### terryds

1. The problem statement, all variables and given/known data

Let p,q,and r be real numbers so that p+q+r = 2 and p2+q2+r2 = 12
Then, the sum of the maximum and minimum value of r is

A. 4/3
B. 2
C. 3
D. 10/3
E. 4

2. Relevant equations
basic algebra

3. The attempt at a solution

p+q+r = 2
p^2+q^2+r^2 + 2 (pq+qr+pr) = 4
12 + 2 (pq+qr+pr) = 4
pq+qr+pr = -4

I think I should determine the value of pqr so that I can treat p,q,r as roots of a polynomial.
But, I have no idea about it.

Last edited: Apr 27, 2016
2. Apr 27, 2016

### Staff: Mentor

Shouldn't the right side of the equation above be 4? It appears that you squared both sides of the first equation.

3. Apr 27, 2016

### terryds

It's fixed (edited :D) now..

4. Apr 27, 2016

### Staff: Mentor

It might help to think about the geometry involved here. The equation p + q + r = 2 is a plane in three dimensions, with intercepts at (2, 0, 0), (0, 2, 0), and (0, 0, 2). The equation $p^2 + q^2 + r^2 = 12$ is a sphere centered at (0, 0, 0), with radius $\sqrt{12}$. The plane and sphere don't intersect in the first octant, but they do intersect in some of the other octants. If you can find any point of intersection, I believe that you will be able to answer the question. I haven't worked the problem, but that's the tack I would take.

5. Apr 27, 2016

### terryds

How to find the intersections? I haven't learned 3d geometry equation yet

6. Apr 27, 2016

### Staff: Mentor

See if this gets you anywhere...
Solve for, say, q, in the first equation (p + q + r = 2).
Substitute for q in the second equation. After substitution you will have the equation of a circle, from which you should be able to find the max and min values of r.

7. Apr 27, 2016

### terryds

q = 2 - r - p

p^2+q^2+r^2 = 12
p^2 + (2-(r+p))^2 + r^2 = 12
p^2 + 4 + r^2+p^2+2rp - 4r- 4p + r^2 = 12
2p^2 + 2r^2 - 4p - 4r - + 2rp - 8 = 0
p^2 + r^2 - 2p - 2r + rp - 4 = 0

Seems like it's not a circle.. since there is rp variable

8. Apr 27, 2016

### Ray Vickson

If you were permitted to use calculus, this would be a multivariable optimization problem with a reasonably straightforward solution using the method of Lagrange multipliers.

If you are not allowed to use calculus, you can do it using algebra only, but it is messy. Solve for p and q in terms of r, using your two given equations. For which values of r do your solutions p = p(r) and q = q(r) make sense?

Last edited: Apr 27, 2016
9. Apr 27, 2016

### Ray Vickson

In (p,q,r)-space the intersection curve is that of a plane intersecting a sphere, so is a circle. However, it is tilted, so when projected down (say to the (p,q)-plane) it will be an ellipse, because when you look at a tilted circle you see an ellipse.

10. Apr 27, 2016

### Samy_A

Never mind. Didn't understand the exercise correctly.

Last edited: Apr 27, 2016
11. Apr 28, 2016

### terryds

How to find the max or min value???

12. Apr 28, 2016

### Ray Vickson

I already suggested a way in post #8: solve for p and q as functions of r, and see what you get. Have you tried that?

13. Apr 28, 2016

### terryds

p as a function of r??
from p^2 + r^2 - 2p - 2r + rp - 4 = 0, how to solve for p?? It's hard since there is rp and it's quadratic..

I know about finding minima and maxima but I don't know what you mean by Langrage multipliers. I just know basic calculus

14. Apr 28, 2016

### Ray Vickson

It is pretty easy to solve the two equations $p+q = S, p^2 + q^2 = T$ for $p, q$ in terms of $S, T$. Once you have done that, put $S = 2-r$, $T = 12-r^2$.
And, yes, indeed, you get a quadratic equation to solve; that is the reason you can get both upper and lower bounds on $r$.

As for a calculus solution, I also told you how to do it in post #8, but unless you are willing and able to use Lagrange multipliers it would not be easy. Google 'Lagrange multipliers' or 'Lagrange multiplier method' for more details, or look in your textbook if you have one.

15. Apr 29, 2016

### terryds

Alright, so q = S - p

$p^2 + (S-p)^2 = T \\ p^2 + S^2 - 2ps + p^2 = T \\ 2p^2 - 2pS = T - S^2 \\ 2p^2 - 2pS + \frac{1}{2} S^2 = T - S^2 + \frac{1}{2}S^2 \\ (\sqrt{2}p-\frac{1}{2}\sqrt{2}S)^2 = T - S^2 + \frac{1}{2}S^2 \\ \sqrt{2}p-\frac{1}{2}\sqrt{2}S = \pm \sqrt{T - S^2 + \frac{1}{2}S^2 } \\ \sqrt{2}p = \frac{1}{2}\sqrt{2}S \frac{}{}\pm \sqrt{T - S^2 + \frac{1}{2}S^2 } \\ p = \frac{\frac{1}{2}\sqrt{2}S \pm \sqrt{T - S^2 + \frac{1}{2}S^2 }}{\sqrt{2}} \\ p = \frac{\frac{1}{2}\sqrt{2}S \pm \sqrt{T - \frac{1}{2}S^2 }}{\sqrt{2}} \\ p = \frac{\frac{1}{2}\sqrt{2}(2-r) \pm \sqrt{(12-r^2) - \frac{1}{2}(2-r)^2 }}{\sqrt{2}}$

Things get messier and messier..
I think I should go learning Langrage Multiplier first.. Thanks..

16. Apr 29, 2016

### Ray Vickson

Why give up now? Do you remember what I said in the last sentence of post #8?

17. Apr 29, 2016

### terryds

$(12-r^2)-\frac{1}{2}(2-r)^2 > 0 \\ (12 - r^2) - \frac{1}{2} (r^2 - 4r + 4) > 0 \\ -\frac{3}{2}r^2 + 2r + 10 > 0$

The roots are 10/3 and -2
The solution is -2 < r < 10/3

The sum is 10/3 - 2 = 4/3

18. Apr 30, 2016

### Ray Vickson

Well, you should write $\geq 0$ rather than $> 0$ and non-strict inequalities $-2 \leq r \leq 10/3$, but other than that you have the basic idea. In other words, the values $r = -2$ and $r = 10/3$ are definitely allowed.

Last edited: Apr 30, 2016