How do I find four positive real numbers whose sum is 2112 and whose product is maximized?

Answer 1

The product of two numbers (whose sum is a constant N) is:   f(x) = x(N-x) = Nx - x²

 f’(x) = N – 2x

The MAXIMUM product occurs when:

      N – 2x = 0

      x = N/2

Now, by substitution, if x is itself the MAXIMUM product of two values, z and (N/2 – z), whose sum is a constant (N/2),    

      f(z) = (N/2)z – z²

the MAXIMUM value of the product occurs when:

       f’(z) = 0 = (N/2) – 2z

             z = N/4

For this problem:

2112 / 4 = 528

528⁴ produces the maximum product of four REAL numbers whose sum is 2112.

Now, if the numbers must be unique, separate them by a small value:

   527.8,      527.9,   528.1,    528.2

       

Answer 2

The sum of numbers are maximized when they are consecutive.

4n + 3 = 2112

4n = 2109

n = 527.25

527.25, 528.25, 529.25, 530.25