The sum of three positive digits is equal to 12. Determine the three digits so that your product is maximum.

Update:

Thank you very much!

Relevance

Suppose one number is a. Then the others add up to 12-a, and they average (12-a)/2

The other two numbers must differ from their average by the same amount, so they are

(12-a)/2 + x

and

(12-a)/2 - x

The product is then a[(12-a)/2 + x][(12-a)/2 - x] = a[ ( (12-a)/2 )^2 - x^2 ]

No matter what a is, this is a maximum when x is 0, so the other two numbers must be equal.

So the product is a X [(12-a)/2]^2

Now you can just look at the cases. You will find that the max occurs when a = 4. I'll do a few cases to illustrate.

a = 1 gives you 1 X 5.5^2 = 30.25

a = 2 gives you 2 x 5^2 = 50

a = 3 gives 3 X 4.5^2 = 3 X 20.25 = 60.75

a = 4 gives 4 X 4^2 = 64

a = 5 gives 5 x 3.5^2 = 5 x 12.25 = 61.25

Keep going up to a = 9 and the products just keep getting smaller.

It doesn't matter that the "other digits" are not integers (e.g, 4.5 or 3.5). Those values are just those that give the maximum, and picking integers would just give you an even smaller product. E.g., if we had digits a=5, others = 3 and 4, the product is 5x3x4 = 60, which is even less than 5 x 3.5^2 = 61.25.

This isn't as sophisticated an answer as others using calculus, but it has the advantage of just using basic algebra, and checking a few easy cases.

If you want to solve it mathematically instead of by checking cases, you can take the product function:

P(a) = 0.25a(12-a)^2 = 0.25(144a - 24a^2 + a^3)

and take the derivative:

P'(a) = 0.25(144 - 48a + 3a^2) = 0.75(48 - 16a + a^2) = 0.75(x-4)(a-12)

That = 0 when a = 4 (the max, other digits are both 4) or a = 12 (the min, other "digits" are both 0) in the range a in [0, 12]. Values of a outside that range are irrelevant.

• Given:

a + b + c = 12

a * b * c = P

Also:

1 ≤ a, b, c ≤ 9

Solving for one variable and substituting:

a = 12 − b − c

P = (12 − b − c) b c

P = 12bc − b²c − bc²

Taking the partial derivative with respect to b and setting to 0:

∂P/∂b = 12c − 2bc − c²

0 = 12c − 2bc − c²

0 = 12 − 2b − c

2b = 12 − c

b = (12 − c) / 2

Taking the partial derivative with respect to c and setting to 0:

∂P/∂c = 12b − b² − 2bc

0 = 12b − b² − 2bc

0 = 12 − b − 2c

2c = 12 − b

c = (12 − b) / 2

We can substitute this equation for c into the previous equation and solve for b, or we can simply deduce that b = c, and through a little trial and error, b = c = 4.

So the three digit number is 444.

• Use the fact that the product of three numbers with a fixed sum is the greatest if the numbers are equal.

So the three numbers are: 4, 4, 4