You can now sell 30 cars per month at $15,000 per car, and demand is increasing at a rate of 2 cars per month each month. What is the fastest you could drop your price before your monthly revenue starts to drop? HINT [Revenue = Price × Quantity.]

Step-by-step explanation:

R = P * Q

Let's make differentiation for time (t)

dR/dt = (dP/dt)*Q + P*(dQ/dt)

dR/dt = (dP/dt)*30 + 15,000*2

The revenue starts to drop means dR/dt < 0

Then:

(dP/dt)*30 + 15,000*2 < 0

(dP/dt)*30 < -30,000

dP/dt < -1,000

Thus, the monthly revenue starts to drop when the price drops more than $1,000 per month.

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺