Answer:
The numbers are;
x = 100/3, y = 100/3, z = 100/3
Step-by-step explanation:
Given the data in the question;
so let the numbers be; x , y and z
so
x + y + z = 100
now
G(x,y,z) = x + y + z - 100
F(x,y,z) = xyz
Fx = yz, Gx = 1
Fy = xz, Gy = 1
Fz = xy, Gz = 1
Fx = λ × Gx
yz = λ
Similarly, xz = lambda
And xy = lambda
From above clearly, x = y = z
x + y + z = 100
so
3x = 100
x = 100/3
since, x = y = z therefore, each number is equal to 100/3
x = 100/3, y = 100/3, z = 100/3
We want to find 3 positive numbers such that their sum is 100 and the product is maximum.
The maximum product is P = 37,025.93
Let's say that our 3 numbers are:
A, B, and C.
We must have:
A + B + C = 100
The product of these 3 numbers is:
P = A*B*C
For the restriction, we can write:
A = 100 - B - C
Now we can replace that in the product equation to get:
P = (100 - B - C)*B*C
P = 100*B*C - C*B^2 - B*C^2
Now we want to maximize this, to do it, we need to find the zeros of the derivate of the function. This is a two-variable function, but the dependence on C is exactly the same as the dependence on B, so we can find the zero of deriving with respect to one of these only.
dP/dB = 100*C - 2*C*B - C^2 = 0
C*(100 - 2*B - C) = 0
One solution is C = 0, the other is:
C = 100 - 2*B
And from the first restroctopm:
A + B + C = 100
we can get:
C = 100 - A - B
Then:
100 - A - B = 100 - 2*B
Then we get A = B
And if we derive the equation for C instead of B, we will get similar results that will lead to:
A = B = C.
This is what maximizes the product, then we have:
A + B + C = 3*A = 100
A = 100/3 = 33.33
P = (33.33)^3 = 37,025.93
This is the maximum product.
If you want to learn more, you can read:
https://brainly.com/question/11212148
