1. ## 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?

2. 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.

3. 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?

4. 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

5. 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.