Question

If x and y are two nonnegative numbers and the sum of twice the first ( x ) and three times the second ( y ) is 60, find x so that the product of the first and cube of the second is a maximum.

Answer 1

2x+3y=60 solve for x...

x=(60-3y)/2

p=xy^3, using x from above:

p=y^3(60-3y)/2

p=(60y^3-3y^4)/2

dp/dy=(180y^2-12y^3)/2

dp/dy=90y^2-6y^2

dp/dy=0 when:

90y^2-6y^3=0

6y^2(15-y)=0, we know y>0 so

y=15, since x=(60-3y)/2

x=7.5

(We solved it this way because otherwise you get quartic equations and cubic derivatives, yuck :P)

Answer 2

2x+3y=60 and f(x,y)=xy^3
Solve for x=30-1.5y
Substitute in
f(y)=(30-1.5y)y^3 = 30y^3-1.5y^4
taking the derivative and finding the critical points yields:
f'(y)=90y^2 -6y^3 = 0
y=15
Then x =30-1.5*15
x=7.5
So x=7.5 and y =15