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October 24, 2014

October 24, 2014

Posted by **Julie** on Saturday, November 24, 2007 at 10:52am.

Can somebody help me answer this and explain the steps to solve the problem?

- Calculus -
**Michael**, Saturday, November 24, 2007 at 11:38amThis is a related rates problem, and all problems of this type have three distinct steps:

(1) Write the general equation for the problem. In this case, we want the equation for revenue.

(2) Implicitly differentiate the equation with respect to time. It's important to note that you should NOT plug in any known values for your variables until AFTER differentiating.

(3) Plug in what you know to solve for what you need.

Try that, and let us know if you get stuck. Good luck.

- Calculus -
**Julie**, Saturday, November 24, 2007 at 11:59amWhat do you mean in step 2?

- Calculus -
**Michael**, Saturday, November 24, 2007 at 12:44pmFor example, if you were given a triangle, your general equation would be the Pythagorean theorem... a^2 + b^2 = c^2.

Find the derivate (by implicit differentiation) for the second step. You'll be using the power and chain rules.

2a (da/dt) + 2b (db/dt) = 2c (dc/dt)

The da/dt is the "rate of change of a with respect to time."

In your problem, your equation will be revenue. Revenue = (# books)(cost of each book). I let b be the number of books and c be the cost of each book.

R = b*c

You have to use the power rule here (first multiplied by the derivate of the second plus second multiplied by the derivate of the first).

dR/dt = (b)(dc/dt) + (c)(db/dt)

Plug in what you know now by analyzing the given information. For example, let's look at this sentence: "Also, it is now selling all its books for $20 each, but the price is dropping at a rate of $1 per week." This gives us the cost of each book (c = 20) and the rate of change of the cost of each book (dc/dt = -1). Do that with all the given information, and solve for the rate of change of revenue (dR/dt).

Try that, and I'll be glad to answer any questions you have.

- Calculus -
**Michael**, Saturday, November 24, 2007 at 1:05pmSorry, the PRODUCT rule is "first * derivate of the second + second * derivate of the first." (I referred to that as the power rule.)

- Calculus -
**Julie**, Saturday, November 24, 2007 at 1:41pmOkay so I got

q=1000+200x

p=20-1x

x=weeks

R=20-1x(1000+200x)

R'=3000-400x

Is this right? If so how do I find how much the revenue is rising or falling and at what rate would sales have to be increasing to accomplish $5000 revenue increase per week?

- Calculus -
**Michael**, Saturday, November 24, 2007 at 2:04pmNo, you're supposed to use the equation...

dR/dt = (b)(dc/dt) + (c)(db/dt)

which is the same as

R' = b(c') + c(b')

We got that from the steps above. Now plug in the information that you know to solve for dR/dt (or R') for the first part.

After you get that, we can work on the second part.

- Calculus -
**Michael**, Saturday, November 24, 2007 at 2:10pmb = the number of books

c = the cost of each book

db/dt = the rate of change of the number of books sold

dc/dt = the rate of change of the cost of each book

We're given all this information in the problem, so all we have to do is plug that into our equation.

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**Julie**, Saturday, November 24, 2007 at 2:27pmI am so confused, this is not making any sense

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**FlOriDa**, Monday, February 4, 2008 at 8:57amwala KayOng naItuLong!!!!!!

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**bobpursley**, Saturday, November 24, 2007 at 3:14pmAt the moment OHaganBooks is selling 1000 books per week

**that is book rate b**and sales are rising at a rate of 200 books per week.that is db/dt Also, it is now selling all its books for $20 each,**P**but the price is dropping at a rate of $1 per week**dP/dt**. I need to know at what rate OHaganBooks' revenue is rising or falling given these conditions**dR/dt**.

++++++++++++++++

Start off with the revenue function.

Revenue= Priceperbook*books/week

dRevenue/dt= Priceperbook* dbrate/week + books/week*dP/dt

or dR/dt= P db/dt + b * dP/dt

and I have highlighted this in the words above. For info, this is not a related rates problem, but just a simple product differentiation.

- Calculus -
**Michael**, Saturday, November 24, 2007 at 3:19pmAll of my work was correct. It can be solved using the same method as a related rates problem. Your equation is the same as the one I came up with.

- Calculus -
**Writeacher**, Saturday, November 24, 2007 at 4:16pmOne thing to keep in mind (if you TRULY want to be a successful teacher) is that it's not the student's job to adapt to the teacher's thinking and explanation. It's the teacher's job to explain concepts in different ways until the student DOES understand.

One thing to do is to pay attention to how seasoned and successful teachers teach, and then emulate them.

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**Destinee**, Sunday, November 25, 2007 at 5:53pmcan someone start over on this problem?

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