The answer given in the textbook is : 16 and 24, however I'm finding two different values, 15 and 25. Any assistance would be greatly appreciated.
The sum of two positive integers is 40. Find the two integers such that the product of the square of one number and the cube of the other number is a maximum:
P = (x^2)(y^3)
As x + y = 40, x = 40 - y
Substituting this back into the product expression:
P = (40-y)^2(y^3)
P = (1600 -80y + y^2)y^3
P = 1600y^3 -80y^4 +y^5
dP/dy = 4800y^2 -320y^3 +5y^4
When such equals zero, 5y^2(960-64y) = 0
Obviously y=0 is not a solution, so 64y = 960 , y = 15
Is there another method of finding the solution?