# optimization problems

• Oct 25th 2009, 04:09 PM
Nightasylum
optimization problems
The sum of the first number and twice the second number is 108 and the product is a maximum.
So I did
x + 2x = 108
x+x=108/2 = 54
2x=54
x=54/2 = 27
54,27 Im not sure if this is the right way to do it?
• Oct 25th 2009, 04:14 PM
skeeter
Quote:

Originally Posted by Nightasylum
The sum of the first number and twice the second number is 108 and the product is a maximum.
So I did
x + 2x = 108
x+x=108/2 = 54
2x=54
x=54/2 = 27
54,27 Im not sure if this is the right way to do it?

you called the second number x also ? did the problem tell you that the first and second numbers were equal?

if not ...

x + 2y = 108

P = xy = (108 - 2y)y

find the value of y that makes P a maximum.
• Oct 25th 2009, 04:29 PM
Nightasylum
Quote:

Originally Posted by skeeter
you called the second number x also ? did the problem tell you that the first and second numbers were equal?

if not ...

x + 2y = 108

P = xy = (108 - 2y)y

find the value of y that makes P a maximum.

Im a little lost on this can you show me step by step?
• Oct 25th 2009, 05:10 PM
Nightasylum
Quote:

Originally Posted by Nightasylum
Im a little lost on this can you show me step by step?

x+2y=108
x=54-y
54-y+2y=108
y=108-54
y=54
2x+54=108
2x=108-54
x=27
So x=27 y=54
Like that?

wait it wont work hmmm
• Oct 25th 2009, 05:19 PM
skeeter
Quote:

Originally Posted by Nightasylum
x+2y=108
x=54-y
54-y+2y=108
y=108-54
y=54
2x+54=108
2x=108-54
x=27
So x=27 y=54
Like that?

wait it wont work hmmm

the product is a maximum ...

$P = (108-2y)y$

you can solve this using calculus ... $\frac{dP}{dy} = 0$ ... and determine the value of $y$ that maximizes $P$

or, without calculus ... you can remember something about the graph of a quadratic function w/ a negative leading coefficient and where the maximum is located.