# Use the mulitplicators of Langrange to find three numbers strictly positive where the sum is 100 and their product is maximum.

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### 1 Answer

You need to randomly select three positive numbers such that: x, y,z.

You know that the sum of the numbers is 100, hence:

`x + y+ z = 100 =gt 3x + 3 = 100 =gt 3x - 97 = 0`

Hence, the function that needs to be optimized is `f(x,y,z) = xyz ` and the constraint is `g(x,y,z)=x + y+ z = 100` .

Hence, you need to solve the equations:

`f_x = lambda*g_x =gt yz = lambda`

`f_y = lambda*g_y =gt xz = lambda`

`f_z = lambda*g_z =gt xy = lambda`

`x + y + z = 100`

You need to set the equations yz and xz equal such that:

`yz = xz`

You may divide by z since `z!=0` such that:

`x = y`

You need to set the equations yz and xy equal such that:

`yz = xy`

You may divide by y since `y!=0` such that:

`x = z`

Since x=z and x=y yields y=z, thus you may substitute x for y and z in constraint `x + y + z = 100` such that:

`x+x+x = 100 =gt 3x = 100 =gt x = 100/3 =gt x = 33.33`

**Hence, evaluating the positive numbers for the product of these numbers to be maximum yields: `x=y=z~~33.33` .**