Question

The demand and cost functions for a product are

p(x) = 110 − 0.01x

C(x) = 30x + 1400

x = # of units produced weekly

If the manufacturer decides to increase production by 70 units per week, find the rate at which profit is changing with respect to time when the weekly production is 3600 units.

Answer 1

thats a profit equation p(x) ?

Answer 2

Yes

Answer 3

Wait no

Answer 4

That's the demand function

Answer 5

then p stands for price

Answer 6

p is price per unit
multiply that by x , number of units produced/sold , to get revenue.
then
profit = revenue - cost

Answer 7

So revenue is R(x) = 110x-0.01x^2

Answer 8

and profit is \[P(x) = 80x - 0.01x^2-1400\]

Answer 9

Now do I differentiate P(x)?

Answer 10

yes, but i got a different function

Answer 11

one moment

Answer 12

sorry you are right

Answer 13

Ok so I got

Answer 14

\[\frac{ dP }{ dt } = 80 - 0.01x(\frac{ dx }{ dt }\]

Answer 15

Oh i did that wrong

Answer 16

So P'(x) = 80-0.02x

Answer 17

i mean

Answer 18

\[P'(x) = 80-0.02x(\frac{ dx }{ dt })\]

Answer 19

$$ \Large {
\Large{

Profit =Revenue - Cost \\
\\ \therefore \\
P(x) = R(x) - C(x) \\
P(x) = x* p(x) - C(x)
\\P(x) = x (110 - 0.01x) - (30x + 1400)
\\P(x) = 80*x-0.01*x^2-1400
\\ \therefore\\
\frac{dP}{dt} = 80\cdot \frac{dx}{dt} - (0.01)\cdot 2x \cdot \frac{dx}{dt}
}

}

$$

Answer 20

After this, what do I do? Plug in 70 for dx/dt and 3600 for x?

Answer 21

Ok I did that and got the answer

Answer 22

is it correct?

Answer 23

Yes :^)

Answer 24

$$ \Large {
\Large{

Profit =Revenue - Cost \\
\\ \therefore \\
P(x) = R(x) - C(x) \\
P(x) = x* p(x) - C(x)
\\P(x) = x (110 - 0.01x) - (30x + 1400)
\\P(x) = 80*x-0.01*x^2-1400
\\ \therefore\\
\frac{dP}{dt} = \frac{dP}{dx} \cdot \frac{dx}{dt} = (80 - 0.01\cdot 2x) \cdot \frac{dx}{dt}
}

}

$$