# Min and max value of r

## Homework Statement

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

basic algebra

## The Attempt at a Solution

[/B]
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:

Mark44
Mentor

## Homework Statement

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

basic algebra

## The Attempt at a Solution

[/B]
p+q+r = 2
p^2+q^2+r^2 + 2 (pq+qr+pr) = 2
Shouldn't the right side of the equation above be 4? It appears that you squared both sides of the first equation.
terryds said:
12 + 2 (pq+qr+pr) = 2
pq+qr+pr = -5

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.

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

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

Mark44
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.

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.

How to find the intersections? I haven't learned 3d geometry equation yet Mark44
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.

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.

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

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

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

basic algebra

## The Attempt at a Solution

[/B]
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.

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:
Ray Vickson
Homework Helper
Dearly Missed
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

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.

Samy_A
Homework Helper
Never mind. Didn't understand the exercise correctly.

Last edited:
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.

How to find the max or min value???

Ray Vickson
Homework Helper
Dearly Missed
How to find the max or min value???

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?

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?

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

Ray Vickson
Homework Helper
Dearly Missed
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

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.

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.

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..

Ray Vickson
Homework Helper
Dearly Missed
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..

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

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

##
(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

Ray Vickson
Homework Helper
Dearly Missed
##
(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

• 