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