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Min and max value of r

  • Thread starter terryds
  • Start date
  • #1
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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

Homework Equations


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.
Please help
 
Last edited:

Answers and Replies

  • #2
33,270
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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

Homework Equations


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.
Please help
 
  • #3
392
13
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..
Please help
 
  • #4
33,270
4,966
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
392
13
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 :cry:
 
  • #6
33,270
4,966
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
392
13
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
 
  • #8
Ray Vickson
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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

Homework Equations


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.
Please help
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:
  • #9
Ray Vickson
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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.
 
  • #10
Samy_A
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Never mind. Didn't understand the exercise correctly.
 
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  • #11
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13
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???
 
  • #12
Ray Vickson
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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?
 
  • #13
392
13
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..

What about using Calculus?
I know about finding minima and maxima but I don't know what you mean by Langrage multipliers. I just know basic calculus
 
  • #14
Ray Vickson
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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..

What about using Calculus?
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.
 
  • #15
392
13
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..
Anyway, what's the answer??
 
  • #16
Ray Vickson
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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..
Anyway, what's the answer??
Why give up now? Do you remember what I said in the last sentence of post #8?
 
  • #17
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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

Thanks for your help
 
  • #18
Ray Vickson
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##
(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

Thanks for your help
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.
 
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