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

[anemone]

If $p,\,q,\,r,\,s$ are positive integers with sum 63, what is the maximum value of $pq+qr+rs$?

 

[RLBrown]

Spoiler

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]

Spoiler

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

Spoiler

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

Spoiler

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

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

 

[anemone]

Spoiler

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

Spoiler

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

Spoiler

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.

edited to provide the solution

Spoiler

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

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:

Spoiler

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.