MHB What Is the Maximum Value of $pq+qr+rs$ When $p+q+r+s=63$?

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The discussion focuses on maximizing the expression \(pq + qr + rs\) under the constraint \(p + q + r + s = 63\) with \(p, q, r, s\) as positive integers. Participants highlight the effectiveness of using geometric shapes, such as rectangles and quadrilaterals, to justify their approaches to the problem. Multiple methods are shared, with one participant providing a solution and another offering an alternative perspective that enhances understanding. The community appreciates the collaborative effort in solving the challenge. Overall, the thread emphasizes the importance of diverse problem-solving techniques in mathematics.
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If $p,\,q,\,r,\,s$ are positive integers with sum 63, what is the maximum value of $pq+qr+rs$?
 
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Two observations:
1) To max rectangular Area for given perimeter. use square. (want squares)
2) Area increases as the square of a side. (want BIG squares)

With some trial and error, I choose
{p, q, r, s} = {1, 2, 30, 30}
for answer of 962
 
RL Brown's starting is right but he overlooked that it need not be a square. A rectangle shall do

p = 1, q = 30, r = 31 , s =1 gives 991

The discussion is informal but if we can raise p and q so that the product is the maximum we can keep r and s as low as possible which gives the value as above

edited to provide the solution

$pq + qr + rs = pq + qr + rs + sp - sp = (p+r)(q+s) - sp$

now we need to maximize $(p+r)(q+s)$ and minimise sp as s and p are in differenent expression

p+r and q + s should be as close as possible as p+r + q + s = 63

so p+ q = 32 and r+ s = 31 or viceversa and p = s = 1

so p = 1 , q = 31, r = 30, s = 1 or p =1, q = 30, r = 31, s = 1
s
 
Last edited:
RLBrown said:
Two observations:
1) To max rectangular Area for given perimeter. use square. (want squares)
2) Area increases as the square of a side. (want BIG squares)

With some trial and error, I choose
{p, q, r, s} = {1, 2, 30, 30}
for answer of 962

kaliprasad said:
RL Brown's starting is right but he overlooked that it need not be a square. A rectangle shall do

p = 1, q = 30, r = 31 , s =1 gives 991

The discussion is informal but if we can raise p and q so that the product is the maximum we can keep r and s as low as possible which gives the value as above

Thanks, RLBrown for participating in this challenge problem. Kaliprasad is right, a rectangle would make the argument more justifiable and thank you so much for giving us another insight to approach this kind of problem using the quadrilateral method.:cool:

kaliprasad said:
edited to provide the solution

$pq + qr + rs = pq + qr + rs + sp - sp = (p+r)(q+s) - sp$

now we need to maximize $(p+r)(q+s)$ and minimise sp as s and p are in differenent expression

p+r and q + s should be as close as possible as p+r + q + s = 63

so p+ q = 32 and r+ s = 31 or viceversa and p = s = 1

so p = 1 , q = 31, r = 30, s = 1 or p =1, q = 30, r = 31, s = 1

Thanks for participating, kaliprasad and your answer is correct!(Smile)

Here is another method that I want to share with the community here that I think you all will find it very illuminating:

For all real $p, \,q$, we have $(p-q)^2\ge 0$. So, $p^2+q^2\ge2pq$, hence $(p+q)^2\ge4pq$ or equivalently, $pq\le\dfrac{(p+q)^2}{4}$.

Letting $x=p+r$ and $y=q+s$ gives $(p+r)(q+s)\le\dfrac{(p+q+r+s)^2}{4}$ so $pq+qr+rs+sp\le\dfrac{63^2}{4}=992.25$.

Since $p,\,q,\,r,\,s$ are positive integers, the last inequality can be written as $pq+qr+rs+sp\le992$. Hence $pq+qr+rs\le992-sp\le991$.

It remains to show that 991 is achievable. Suppose $pq+qr+rs=991$ and $p=s=1$, then $(1+q)(1+r)=992=2^5\cdot31$. So $q=30$ and $r=31$ is a solution. Thus the maximum of $pq+qr+rs$ is 991.
 
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