Question

A manufacturer produces bolts of a fabric with a fixed width. The quantity of this fabric (measured in yards) that is sold is a function of the selling price p (in dollars per yard), so we can write q = f(p). Then the total revenue earned with selling price p is R(p) = pf(p).

a. What does it mean to say that f(20) = 10,000 and f'(20) = -350?
b. Assuming the values in part (a), find R'(20) and interpret your result

Answer 1

@zepdrix can you please help me with this question?

Answer 2

Hmm interesting problem :)
Thinking..

Answer 3

oh okay...
good luck....

Answer 4

Let's try to interpret the f(20) first.
Recall that this function is defined to be our `quantity`.
And the function depends on the `price`.

So at a price of p=20, we are able to sell 10000 bolts of fabric.
f(20)=10000.

Answer 5

okay....
I get that now

Answer 6

The f'(20) is a little trickier.

The derivative represents the instantaneous rate of change of the function,
the quantity function,
at p=20.

So at a price of p=20,
The amount of bolts of fabric we are selling
is decreasing by 350 bolts per... dollar maybe?

Answer 7

That should be it

Answer 8

it makes a lot of sense to put it that way

Answer 9

Oh they gave us the units at the start :)
I forgot about that.

So our at a price of $20,
Our quantity is decreasing by 350 yards of fabric per dollar.

Ya ya ya, I think that makes sense.

Answer 10

For part b,
\[\large\rm R(p)=p~f(p)\]To get the derivative of the revenue function,
you'll need to apply your product rule.
Remember that one? :)

Answer 11

I understand that....
i really don't know how to start

Answer 12

p is not a function, it is a constant.
is it necessary to use the product rule?

Answer 13

I guess it's not a constant,
p is price.

Price can range from any number of values that we set it at.
So price is a variable.
It `will be` constant, when we set it at $20.
But we have to apply our differentiation rule `before` we do that.

Answer 14

What we're doing is, looking for a derivative function `with respect to p`.\[\large\rm R'=\frac{dR}{dp}\]

Answer 15

oh okay....
i get it

Answer 16

Applying Product Rule gives us this "set up",\[\large\rm R'(p)=p'~f(p)+p~f'(p)\]That rule looks familiar, yes? :)

and hopefully you can make sense of the way p' will simplify.
It's just the derivative of p with respect to p.
Similar to if you had taken a derivative like this,\[\large\rm \frac{d}{dx}x=1\]

Answer 17

So our Derivative Revenue Function looks like this,\[\large\rm R'(p)=1\cdot f(p)+p~f'(p)\]\[\large\rm R'(p)=f(p)+p~f'(p)\]

Answer 18

\[\large\rm R'(\color{orangered}{p})=f(\color{orangered}{p})+\color{orangered}{p}~f'(\color{orangered}{p})\]And we want to evaluate this derivative at p=20,\[\large\rm R'(\color{orangered}{20})=f(\color{orangered}{20})+\color{orangered}{20}~f'(\color{orangered}{20})\]

Answer 19

To do so,
you'll need to use the `pieces of information they gave you` from before:\[\large\rm f(20)=10000,\qquad\qquad f'(20)=-350\]

Answer 20

Thank you very much
I get it now
I got my answer to be 3000 dollars

Answer 21

Yay good job \c:/
np