Question

Two positive numbers have a sum of 60. what is the maximum product of one number times the square of the second number?
a) 3481
b) 3600
c) 27,000
d) 32,000
e) 36,000

So Idk if this is right, but I have ....
x+y=60
p=xy^2
p=x(60-x)^2
P'=-2x(60-x)+(60-x^2)
x=.5021 x= 119.498
idk if this is right so far, or if i forgot something. how do I find the max product?

Answer 1

why did you square the more complicated factor, why not square the x

p = x^2(60-x)
= 60x^2 - x^3

dp/dx = 120x - 3x^2
= 0 for a max of p
3x(40 - x) = 0
x = 0 or x = 40
obviously x=0 will produce the minimum product

if x = 40
then 60-40 = 20

so the two numbers are 40 and 20 (with 40 as the number that was squared in the product)

Your way should have worked too, but I notice that you jumped from
.... (60-x)^2 to
.....(60-x^2) , that is an error.

my product is (40^2)(20) = 32 000

btw, your two numbers don't even add up to 60