Find three positive numbers x y and z that satisfy the given

Solution

Sum of x,y,z=56
Solving step by step
z = 56-x-y
P = xy^2z
P = xy^2(56-x-y)
P = 56xy^2 - x^2y^2 - xy^3
dP/dx = 56y^2 - 2xy^2 - y^3
dP/dy = 112xy - 2x^2y - 3xy^2
dP/dx = 0 gives
56y^2 - 2xy^2 - y^3 = 0
y^2(56 - 2x - y) = 0
2x+y = 56.........(1)
dP/dy = 0 gives
112xy - 2x^2y - 3xy^2 = 0
xy(112 - 2x - 3y) = 0
2x + 3y = 112.......(2)
solving 1 & 2, applying (1) - (2)
-2y = -56
y = 28
putting it in (1)
2x + 28 = 56
2x = 28
x = 14
x+y+z = 56
14+28+z = 56
z = 14
therefore x = 14, y = 28, z = 14