# Max and Min Problem

1. Apr 24, 2005

### scorpa

Hello Everyone,

I am having trouble with a maximum problem and I'm not quite sure where I am going wrong so I will type of the problem and what I have done so far.

Find two positive numbers whose sum is 18 and the product of the first number and the square of the other is a maximum.

Here is what I've done so far:

x + y = 18 ---> y = 18 - x

xy^2=P

P = x(18-x)^2
P = x(324 - 36x + x^2)
P= x^3 - 36x^2 + 324X

To find where there is a maximum I found the first derivative of the equation above:

dP/dx = 3x^2 - 72x + 324

This is where I'm stuck, I know I want to make the first derivative equal to zero so I can find the values for the maximum, and verify my answer using the second derivative, but the first derivative cannot be factored. I must be doing something terribly wrong. The first time I did it I got x = 18 and y = 0 which cannot be right, to get that answer I took out a common x value in the equation P equation, but when I did the second derivative test it showed that the answer was actually a minimum. If anyone could give me some direction here I would really appreciate it.

2. Apr 24, 2005

### whozum

It doesnt need to be factorable, worst come to worst, you can use the quadratic formula.

Also dP/dx does simplify.

$$(18-x)^2 - 2x(18-x) = (18-x)(18-x-2x) = (18-x)(18-3x)$$

3. Apr 24, 2005

### scorpa

Ok turns out I'm mighty dumb...lol.

so dP/dx= (x-18)(x-6) = 0
therefore x can equal 18 or 6
when you take the second derivative the only number that equals a maximum is 6, so the two numbers are x = 6 and y = 12.

I think that is right now.

4. Apr 24, 2005

### whozum

This doesnt equal $3x^2 - 72x + 324$

5. Apr 25, 2005

### dfollett76

Actually it does, since 3, 72, and 324 are all divisible by three the polynomial can be simplified to x^2 -24x + 108 which can be factored to be (x -18)(x -6)

6. Apr 25, 2005

### whozum

But you still have to tag a 3 on there.. expand your factors.. you get x^2-24x+108, which is 3 times less than the original polynomial.

When you solve for x you can eliminate it though, which is what I think ur doing.

7. Apr 29, 2005

### dfollett76

Oh right, my bad. Just the sort of thing I'd get on a students case about writing, I'm such a hypocrite. :(