Tutor.com The Derivative Session

May. 26, 2012

Session Transcript - Math - Calculus, 5/20/2012 7:56PM - Tutor.comSession Date: 5/20/2012 7:56PM
Length: 34 minute(s)
Subject: Math - Calculus

System Message
[00:00:00] *** Please note: All sessions are recorded for quality control. ***

Guest (Customer)
[00:00:00] We wish to construct an open topped box with a square base that has a volume of 864 cubic inches. Find the dimensions we should use if we want to minimize the amount of cardboard material to be used.

Charles C (Tutor)
[00:00:10] Welcome to Tutor.com ! My name is Charlie.

Guest (Customer)
[00:00:18] hey charlie

Charles C (Tutor)
[00:00:42] So I see you've got a word problem to work on. Did you have any work done on it yet?

Guest (Customer)
[00:00:48] some what
[00:00:51] not much
[00:00:56] all i know is this

Charles C (Tutor)
[00:01:00] OK.

Guest (Customer)
[00:01:14] now
[00:01:21] im thinking that i need to find an equation for the area
[00:01:28] but i can't seem to think of it

Charles C (Tutor)
[00:01:43] OK. That's a good start.
[00:02:21] So you're right, we need to get an expression for the area and then minimize that...
[00:03:10] One thing we can notice before we start that is, given that the box has a square base, what can we say about the relationship of l and w here?

Guest (Customer)
[00:03:09] i think i know one

Charles C (Tutor)
[00:03:13] OK.

Guest (Customer)
[00:03:21] here's what i think it is

Charles C (Tutor)
[00:03:27] OK...
[00:04:40] OK! That looks pretty close...
[00:05:12] Just notice that it's an open-topped box, so how would you modify that formula for a box without a top?

Guest (Customer)
[00:05:21] oh ok

Charles C (Tutor)
[00:05:45] Good!
[00:05:52] We're making progess.
[00:06:16] Now you've got a formula for the quantity you want to minimize...

Guest (Customer)
[00:06:25] so solve for one variable in terms of another?
[00:06:33] in terms of the other two

Charles C (Tutor)
[00:07:10] That's the right idea, but we can actually get it down to 2 variables by looking at all the info in the problem.

Guest (Customer)
[00:07:20] ok

Charles C (Tutor)
[00:07:37] Notice that the box has a square base, so what does that tell us about l and w?

Guest (Customer)
[00:07:45] sameEnter your text or URL here and press Enter to send.
[00:07:47] same

Charles C (Tutor)
[00:07:53] Correct!
[00:08:21] So we can replace all the l's by w's (or vice versa, whichever you prefer).

Guest (Customer)
[00:08:27] ok

Charles C (Tutor)
[00:09:09] Then we can use the idea about solving for one variable in terms of another.

Guest (Customer)
[00:09:16] ok

Charles C (Tutor)
[00:09:51] So we want our A (area) to be a function of just one variable so we can minimize it. I see you're working on that....

Guest (Customer)
[00:10:11] ok now let me plug that into the area equation and simplify it
[00:10:17] i'll get back to you in a minute

Charles C (Tutor)
[00:10:29] OK. I'll work on it as well....

Guest (Customer)
[00:10:31] k
[00:11:07] got it

Charles C (Tutor)
[00:11:18] OK. What did you arrive at?
[00:12:44] OK. We're writing A as a function of h...

Guest (Customer)
[00:13:41] so find the derivative of this?

Charles C (Tutor)
[00:13:50] We had A=4wh+w^2. I think one of the 2wh terms got dropped...

Guest (Customer)
[00:14:02] oh
[00:14:03] ok

Charles C (Tutor)
[00:14:32] And the w^2 still has an h in the denominator.

Guest (Customer)
[00:15:12] ok
[00:15:40] now find the derivative?

Charles C (Tutor)
[00:15:50] But yes, we find the derivative of this. Let's see how the calculations work out. I might be tempted to solve for h in terms of w and have A be a function of w to avoid the square root, but this should work....
[00:15:58] Yes, let's take the derivative.

Guest (Customer)
[00:16:02] ok
[00:16:10] get back to you in a minute again

Charles C (Tutor)
[00:16:26] OK. I'll work on it too...

Guest (Customer)
[00:16:53] go it
[00:16:55] got it

Charles C (Tutor)
[00:17:07] All right! What did you get?

Guest (Customer)
[00:17:19] this

Charles C (Tutor)
[00:18:24] Let me look at that a second...

Guest (Customer)
[00:18:29] okay

Charles C (Tutor)
[00:19:20] I got something a little different on the first term...

Guest (Customer)
[00:21:10] i think i forgot to apply the chain rule on the first one

Charles C (Tutor)
[00:22:16] Does that look all right to you?

Guest (Customer)
[00:22:19] yes

Charles C (Tutor)
[00:22:27] OK. So what comes next?

Guest (Customer)
[00:22:34] set it equal to zero
[00:22:35] 0
[00:22:38] and solve for h

Charles C (Tutor)
[00:22:43] Good!
[00:22:55] Let's see how messy the algebra gets...

Guest (Customer)
[00:22:58] haha

Charles C (Tutor)
[00:23:31] Do you want to try to work that out on paper again?

Guest (Customer)
[00:23:46] we can use a calculator if possible
[00:23:51] if not then we can work it out

Charles C (Tutor)
[00:24:22] OK. I'm just seeing how the calculations work out. Give me just a minute...

Guest (Customer)
[00:24:32] ok
[00:24:38] im confused on how to solve for h

Charles C (Tutor)
[00:25:09] I think it's looking okay. Just a couple more seconds...

Guest (Customer)
[00:25:25] ok

Charles C (Tutor)
[00:26:55] I got something for h that looks a little ugly, but probably okay for the calculator.
[00:27:13] Let's see what happens solving for h here.

Guest (Customer)
[00:27:16] ok

Charles C (Tutor)
[00:28:35] It seems like a good idea to get rid of the fractions by multiplying both sides of the equation by h^2. Sound reasonable?

Guest (Customer)
[00:28:41] yes

Charles C (Tutor)
[00:28:46] OK...

Guest (Customer)
[00:30:06] so h ^3/2 = 14.6969

Charles C (Tutor)
[00:30:36] Looks right. And so h=?

Guest (Customer)
[00:30:56] i got h to equal 3.83

Charles C (Tutor)
[00:31:37] I got h very close to 6. Let me double-check...

Guest (Customer)
[00:31:42] oh.
[00:32:27] yes