Max it out

Let $a, b$ and $c$ denote the number of -1's, 1's and 2's in the sequence, respectively. We choose these numbers because they are between -1 and 2. We do not need to consider the zeros. This means that $a, b$ and $c$ are nonnegative integers satisfying

$-a + b + 2c = 19$ and $a + b + 4c = 99$

Subtracting the equations, you get $2a + 2c = 80$ and $a = 40 - c$ . By adding the equations, you get $2b + 6c = 118$ and $b = 59 - 3c$ .

It also follows that, since $b \geq 0$ , the value of $c$ ranges between 0 and 19.

$x_1^3 + x_2^3 + \ldots + x_n^3 = -a + b + 8c = 19 + 6c$

Since we need to focus only on the maximum value, we will input $c$ as 19 into the equation. This results in the answer being $133$ .

A sum equal to 2 can be obtained using positive numbers in either of the following cases: {2} ==> 2^3 = 8
{1, 1} ==> 1^3 + 1^3 = 2 The first observation is that {2} is more optimal than {1, 1} as the former contributes more towards the sum of the cubes (because 8 > 2). Therefore, we can claim that the sum of 19 would be obtained using nine 2's and one 1's and the sum of the remaining of the numbers would be 0. Sum of squares of these 10 numbers = 9 (2^2) + 1 (1^2) = 37.

A zero sum can always be maintained by any of the following combinations in the boundary cases: {0}, {-1, 1}, {-1, -1, 2} Sum of squares for the 1st case = 0^2 = 0; Sum of cubes for the 1st case = 0^3 Sum of squares for the 2nd case = (-1)^2 + 1^2 = 2; Sum of cubes for the 2nd case = (-1)^3 + 1^3 = 0 Sum of squares for the 3rd case = (-1)^2 + (-1)^2 + 2^2 = 6; Sum of cubes for the 3rd case = (-1)^3 + (-1)^3 + 2^3 = 8

From the above data, it can be said that the preference towards selecting for remaining numbers which contribute a zero sum would in the order of {-1, -1, 2}, {-1, 1}, and {0}. Equating for the sum of squares to be = 99 - 37 = 62 = 10 6 + 1 2, we would select ten {-1, -1, 2} and one {-1, 1}. All the remaining numbers would all be equal to 0.

Thus, the best possible way to maximize the sum of the cubes, we would have: Nineteen 2's, Two 1's, and Twenty-one -1's

Maximum{x(1)^3 + x(2)^3 + x(3)^3 + ... + x(n)^3} = 19 (2^3) + 2 (1^3) + 21 ((-1)^3) + (n - 42) (0^3) = 152 + 2 - 21 + 0 = 133

only possible values of x are -1, 1 ,2 (0 does not affect addition). Lets have a -1's, b1's c2's so -a+b+2c = 19, a+b+4c=99 and find max of -a+b+8c the 2 eqn gives -a+b+8c = 137-2b So we have to find min of b that satisfies 2 equation keeping in mind a,b,c are +ve integers. Now b=0 and b=1 dont give any solution, but b = 2 does...so 137-2b = 133 is the answer.

Let n3, n0, n1, and n2 be the number of -1, 0, 1, 2, respectively. Solve the equations -n3 + n1 + 2 n2 = 19, n3 + n1 + 4 n2 = 99 for n3 and n1 in terms of n2, yielding: n3 = 40 - n2; n1 = 59 - 3 n2, which must be greater or equal to zero, thence n2<20. The sum of cubes S3 = -n3 + n1 + 8 n2 = 19 + 6 n2 is maximum for the greatest n2 allowed, which is 19. Thence S3 = 19 + 6 19 = 133.