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

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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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