• Support PF! Buy your school textbooks, materials and every day products Here!

Calculus Word Question

  • Thread starter Jabus
  • Start date
2
0
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

http://www.jabussucks.com/calcanswer.jpg [Broken]
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 
Last edited by a moderator:

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
41,732
893
At one point you have
[tex]P= 2x+ 3\left(\frac{1200}{x}[/tex]
which is correct. The next line is
[tex]= 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}[/tex]
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
[tex]P= \frac{5400x^2+1}{5400x}[/tex]
which is incorrect. I only mention it because you then have
[tex]P'= \frac{10800x}{5400}[/tex]
No, no, no!! The derivative of [itex]\frac{f}{g}[/itex] is NOT [itex]\frac{f'}{g'}[/itex]!
As if that weren't bad enough, from
[tex]\frac{10800x}{5400}= 0[/tex]
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.
 
2
0
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.
 
Last edited by a moderator:
cristo
Staff Emeritus
Science Advisor
8,056
72
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

[tex]x^{-2}=\frac{2}{3600}[/tex]
Note that [tex]\frac{2}{3600}=\frac{1}{1800}[/tex]. 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 [itex](5.56 \times 10^{-4})^{-1/2}[/itex]
 
HallsofIvy
Science Advisor
Homework Helper
41,732
893
Just seconding what cristo said: you have 3600x-2- 2= 0 so
3600x-2= 2 or x-2= 2/3600= 1/1800. That means that x2= 1800= (2)9(100). [itex]x= \pmsqrt{2}(3)(10)= \pm30\sqrt{2}[/itex]. That's approximately 42.4. (Since x is a distance, it must be positive.)
 

Related Threads for: Calculus Word Question

  • Last Post
Replies
1
Views
825
  • Last Post
Replies
3
Views
708
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
14K
Top