What Is the Sum of the Maximum and Minimum Values of r Given These Conditions?

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In summary: 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... like quadratic equation for pIt's a quadratic equation in p, you need to solve for p in terms of r. Use the quadratic formula. You can factor out r from the p^2 and -2rp terms to make it look a little nicer first though.
  • #1
terryds
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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:
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  • #2
terryds said:

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
Mark44 said:
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
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
Mark44 said:
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
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
Mark44 said:
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
terryds said:

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?
 
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  • #9
terryds said:
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
Never mind. Didn't understand the exercise correctly.
 
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  • #11
Ray Vickson said:
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
terryds said:
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
Ray Vickson said:
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
terryds said:
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
Ray Vickson said:
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
terryds said:
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?
 
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  • #17
Ray Vickson said:
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
terryds said:
##
(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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What is the significance of finding the min and max value of r?

Finding the min and max value of r is important because it helps us understand the strength and direction of the relationship between two variables. It allows us to determine how closely the data is clustered around the best-fit line and if there are any outliers that may be affecting the correlation.

How do you calculate the min and max value of r?

The min and max value of r can be calculated using statistical software or by hand using the formula r = (nΣxy - ΣxΣy) / √((nΣx^2 - (Σx)^2)(nΣy^2 - (Σy)^2)). This formula takes into account the sample size (n), the sum of the products of the x and y variables (Σxy), the sum of the x values (Σx), and the sum of the y values (Σy).

What do the min and max values of r tell us about the relationship between two variables?

The min and max values of r tell us the range of possible correlation coefficients for the given data. A value of r close to -1 indicates a strong negative correlation, a value close to 0 indicates no correlation, and a value close to 1 indicates a strong positive correlation.

Can the min and max values of r change?

Yes, the min and max values of r can change depending on the data being analyzed. Different datasets may have different ranges of possible correlation coefficients, so the min and max values of r may vary.

How do outliers affect the min and max values of r?

Outliers can have a significant impact on the min and max values of r. If there are extreme values in the dataset, they can pull the correlation coefficient closer to -1 or 1, resulting in a larger range of possible values for r. It is important to identify and address outliers when calculating the min and max values of r to ensure an accurate representation of the relationship between the variables.

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