Answer: 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.
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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