# Calculus Word Question

Okay, the following is just me trying to confirm if I did the problem correctly. It seems like I made a mistake somewhere, but the answer does sort of fit. So I suppose I'm just look for a check to see if I did this correctly. The equations I got from the world problem are top right, I then move into the problem on the left side and finish on the right. :D

1. Homework Statement

http://www.jabussucks.com/calcquestion.jpg [Broken]

3. The Attempt at a Solution

1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

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HallsofIvy
Homework Helper
At one point you have
$$P= 2x+ 3\left(\frac{1200}{x}$$
which is correct. The next line is
$$= 2x+ \frac{3600}{3x}[/itex] which is wrong: to multiply a fraction by 3, you multiply the numerator, not both numerator and denominator! = 2x+ 3600*3x-1 Again wrong: you divide by the 3. But since that 3 is wrong to begin with, you've just made it worse. What you should have is P= 2x+ 3600 x-1. Next you have = x+ 5400x-1 that "5400" is your 10800 divided by 2. Apparently you have divided through by 2. But you have to divide both sides of the equation by 2. This would be P/2, not P. Anyway, it's best to leave it as P= 2x+ 3600x-1 and differentiate that. Next, you have [tex]= x+ \frac{1}{5400}$$
No, the "-1" exponent applies only to x, not the "5400". There is, in any case, no point in moving the 1/x up to x-1 only to put it back in the denominator. Go ahead and differentiate P= 2x+ 3600x-1.

Unfortunately, having got the incorrect fraction, you add the fractions to get
$$P= \frac{5400x^2+1}{5400x}$$
which is incorrect. I only mention it because you then have
$$P'= \frac{10800x}{5400}$$
No, no, no!! The derivative of $\frac{f}{g}$ is NOT $\frac{f'}{g'}$!
As if that weren't bad enough, from
$$\frac{10800x}{5400}= 0$$
you go to
10800x= 5400
and solve that to get x= 1/2. No! 5400*0= 0 not 5400!

Better is to take P= 2x+ 3600x-1 and differentiate it directly.
P'= 2- 3600x-2= 0. Then x2= 1800. Solve that for x.

Thanks for the reply HallsofIvy, I just wanted to check in on something. All of your corrections make sense, and clearly my brain isn't functioning today because those are all mostly things I should have caught myself doing.

But here's what happens when I try to go forward with P' = 2 - 3600x^-2

http://www.jabussucks.com/calcfollowup.jpg [Broken]

So from there sub 0 for the P'. At which poin I can move the -3600x^-2 over to the left which gives me 3600x^-2 = 2

To isolate x do I not divide by 3600 both sides and then take the -2root of the right side? Perhaps, I'm just looking at everything wrong, I'm not sure what it is that's throwing me off.

I know If I go with your equation above:
x^2= 1800
then x = 42.4 but I'm hoping to find out what I'm doing wrong that I can't even get to that point.

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cristo
Staff Emeritus
You are at that point! When you work out your value for x it will be 42.4! But, to do it the way that HoI did (and to keep things as fractions, which is always more a more exact way of calculating things) you have the line

$$x^{-2}=\frac{2}{3600}$$
Note that $$\frac{2}{3600}=\frac{1}{1800}$$. Now, take the reciprocal of both sides, and you will obtain x2=1800.

A quick comment about your notation in the penultimate line: The right hand side would generally be written as $(5.56 \times 10^{-4})^{-1/2}$

HallsofIvy
3600x-2= 2 or x-2= 2/3600= 1/1800. That means that x2= 1800= (2)9(100). $x= \pmsqrt{2}(3)(10)= \pm30\sqrt{2}$. That's approximately 42.4. (Since x is a distance, it must be positive.)