The sum of three non-negative numbers is 36, and one of the numbers is twice one of the other numbers. What is the maximum value of the product of these three numbers?

The maximum value of the product of the three given numbers is;

P_max  = 726

Let the three non-negative numbers be x, y and z

We are told that they sum up to 36.

Thus;

x + y + z = 36.

Now, one of the numbers is twice one of the other numbers.

Thus;

x = 2y

Then, we now have;

2y + y + z = 36

⇒ 3y + z = 36.

We want to find the maximum value of the  product between the 3 numbers. Let's say their product is P, then we have;

P = xyz

P = 2y²z

Since y is squared here, then let's make 3y the subject in 3y + z = 36.

3y + z = 36.

3y = 36 - z

since 3 is multiplied by y, it means that  36 - z be a multiple of 3 and we can infer that z must also be a multiple of 3. If we take z = 3, we have;

Then if we have:

3y = 36 - 3

3y = 33

y = 33/3

y = 11.

Since Product is P = 2y²z, then we have;

P = 2 × 11² × 3

P_max  = 726

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726.

Step-by-step explanation:

we have:

a, b and c non-negative numbers.

we know that:

a + b + c = 36.

(i also assume that the numbers are whole numbers)

and "one of the numbers is twice one of the others"

Then we can write:

a = 2*b

replacing that in the above equation, we get:

2*b + b + c = 36

3*b + c = 36.

Then the product between the 3 numbers will be:

P = a*b*c = (2*b)*b*c = 2*b^2*c.

Because we have b squared in the product, we want to find the maximum possible value of b, such that:

3*b + c = 36.

3*b = 36 - c

We know that 36 is a multiple of 3, and also is 3*b.

Then 36 - c must also be a multiple of 3, this means that c must be a multiple of 3 (and the smallest possible), which is 3.

Then if we have:

3*b = 36 - 3 = 33

3*b = 33

b = 33/3 = 11.

The prodct will be:

P = 2*(11^2)*3 = 726.

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