# Optimisation problem help

• Sep 9th 2012, 03:45 AM
DonGorgon
Optimisation problem help
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?
• Sep 9th 2012, 04:11 AM
BobP
Re: Optimisation problem help
Look again at the line where you remove $\displaystyle 5y^{2}$ as a factor.
• Sep 9th 2012, 07:45 AM
MarkFL
Re: Optimisation problem help
As indicated, you have factored incorrectly, but to answer your question about whether there is another method, one can use Lagrange multipliers. We are given the function:

$\displaystyle P(x,y)=x^2y^3$

subject to the constraint:

$\displaystyle g(x,y)=x+y-40=0$

giving the system:

$\displaystyle 2xy^3=\lambda$

$\displaystyle 3x^2y^2=\lambda$

which implies:

$\displaystyle 3x^2y^2-2xy^3=0$

$\displaystyle xy^2\left(\3x-2y\right)=0$

The first factor gives us the possible solutions:

$\displaystyle (x,y)=(0,40),\,(40,0)$ both of which yield:

$\displaystyle P(x,y)=0$

The second factor gives:

$\displaystyle x=\frac{2}{3}y$

and substituting into the constraint, we find:

$\displaystyle \frac{5}{3}y=40\:\therefore\:y=24,\,x=16$