Tennis is Complex!

7^2 + 6^2 + 5^2 +4^2 +3^2 + 2^2+ 1^2 + 0^2= 140

There are 28 matches and eight players.

Each person's number of games won is squared to find the amount of galleons they won.

This can be represented as:

a^2 + b^2 + c^2 + d^2 + e^2 + f^2 +g^2 + h^2= total amount of winnings, which we want to maximize

Each variable stands for the number of games that each person won.

a, b, c , d, e, f, g and h are each less than or equal to 7. They are all integers.

a+b+c+d+e+f+g+h=28 as 28 games were won in all, since 28 games were played

To maximize this equation, one might think that we should start out with each of the first seven variables = 4 and the last one = 0. However, it is difficult to prove/ disprove this, so let's simplify the question.

If we had two variables, x and y, whose sum was 7, what is the maximum value you can get for x^2+y^2?

If we tried x and y =4, we would get 32.

If we increased x by 1 and decreased y by 1, we see that the increase far outweighs the decrease.

When x=5 and y=3, the problem would yield 34.

We quickly see that the maximum occurs when x=7 and y=0.

So does that mean that we should have a,b,c and d= 7 and e,f,g and h =0?

We get 7^2 + 7^2 + 7^2 +7^2 +0^2 +0^2 + 0^2 + 0^2= 140

It seems correct mathematically. All numbers are each less than or equal to 7. They all add up to 28.

But no, this would not be possible, as person a has to play every single person. It would not be possible for him to win seven games and for person b to win seven games, as that means they both won the game against each other, which would make no sense. We need to pick a loser in this match. Let's say that b lost. Therefore, b should be decreased by 1, to 6. But this means that g should be increased by 1, so that all the variables still add up to 28.

We have: 7^2 + 6^2 + 7^2 +7^2 +0^2 +0^2 + 1^2 + 0^2= 140

Now, person c could not have won seven matches, as person a won all of his matches and person b only lost one match, which was to person a. Therefore, we know person c lost two of his matches. We decrease the value of c to 5. This means we have to increase f to 2, like before.

We have: 7^2 + 6^2 + 5^2 +7^2 +0^2 + 2^2 + 1^2 + 0^2= 140

By the same reasoning, d becomes 4 and e has to become 3.

We have 7^2 + 6^2 + 5^2 +4^2 +3^2 + 2^2 + 1^2 + 0^2= 140

Now, our answer is possible and mathematically correct. This provides the maximum value possible: 140.

It was easy !!!!!!!! Just had to multiply all sets possible with 5 i.e. 28*5=140

Follow-up: what's the minimum total payout?

The solution revolves around maximizing the no. of wins so that we arrive at the biggest possible sum of squares. That is only possible, if we have the highest numbers to square and add.

The highest possible number of wins for a player is 7, which means no other player can have 7 wins. The next highest possible number of wins will be 6, and no other player except the first can have the same wins. We go on until we reach 0 wins for the last player. In this process, we have to take care of the total number of wins, which is 28, and make sure our choice of number of wins for all the players add up to it.

So, the solution would be (7^2 + 6^2 + 5^2 + 4^2 + 3^2 + 2^2 + 1^2) = 140

To support this answer, we can take up another possibility for the number of wins. Lets assume all the players have 4 wins and 3 losses, except the last player, who has no wins at all. Then we get (4*7 = 28) wins in total. Now when we square all those wins and add them up, we get the sum as 112, and not 140. So, maximizing the equation is the concept for this problem.

There are 28 sets in total. And as per the ques. each single set is won by any of the two players. Therefore the federation has to anyway give 5 galleons to the winner in each match. Hence the answer,

$28 \times 5 = 140$

As each set results in a victory for one player or another, the federation must quite surely pay 5 galleons for each of the 28 sets. So the answer is 28x5=140.