# Finding certain real numbers...

• September 11th 2009, 08:53 PM
BeSweeet
Finding certain real numbers...
I need to find two positive real numbers whose product is a maximum and where the sum of the first and three times the second is 60.

And apparently, there're 5 steps to doing this, and I know none of them:
http://img17.imageshack.us/img17/334...90911at953.png
• September 11th 2009, 09:10 PM
tutor.mathitutor
step 1
1.x + 3. y = 60

step 2
y = (60-x)/3

step 3
p(x) = x((60-x)/3)

step 4
p(x) = 20.x -(x^2)/3
step 5
max(p(x)) = 300
occurs when x = 30 and y = 10
• September 11th 2009, 09:10 PM
VonNemo19
Sweet, understand what the question is saying....

The Product $P$ is given by $xy=P$. The sum is given by ... what?
• September 11th 2009, 09:14 PM
BeSweeet
Well, it could be by a lot of things... 60*1, 30*2, 6*10...
• September 11th 2009, 09:15 PM
BeSweeet
Quote:

Originally Posted by tutor.mathitutor
step 1
1.x + 3. y = 60

step 2
y = (60-x)/3

step 3
p(x) = x((60-x)/3)

step 4
p(x) = 20.x -(x^2)/3
step 5
max(p(x)) = 300
occurs when x = 30 and y = 10

You are amazing! Although step 4 didn't work right. I'll be damned if I know how you did that.
• September 11th 2009, 09:19 PM
VonNemo19
Quote:

Originally Posted by BeSweeet
Well, it could be by a lot of things...

No...it can't.

It given by "the sum of the first and three times the second"

$\overbrace{x}^{\text{First}}\underbrace{+}_{\text{ sum}}\overbrace{3y}^{\text{3 times the second}}$

$x+3y=60$