P(x) Q(x) = R(x); if P(x) 72 Am R(x); if P(x)=x+2 and R(x) = x³ - 3x²

A scuba diving company currently charges $100per dive. On average, there are 30 customers per day. The company performed a study

Semenov [28]

A. Total Revenue (R) is equal to price per dive (P) multiplied by number of customers (C). We can write $R=PC$ .

Per price increase is $20. So four price increase is $ $20*4=$80$ . Hence, price per dive is 100+80=$180.

Also per price increase, 2 customers are reduced from 30. For 4 price increases, $4*2=8$ customers are reduced. Hence, total customers is $30-8=22$ .

So Total Revenue is:

$R=180*22=3960$

B. Each price increase is 20. So x price increase is 20x. Hence, new price per dive would be equal to the sum of 100 and 20x.

Also per price increase, customers decrease by 2. So per x price increases, the customer decrease is 2x. Hence, new number of customers is the difference of 30 and 2x.

Therefor we can write the quadratic equation for total revenue as the new price times the new number of customers.

$R=(100+2x)(30-2x)=-40x^{2}+400x+3000$

C. We are looking for the point (x) at which the equation modeled in part (B) gives a maximum value of revenue (y). That x value is given as $x=-\frac{b}{2a}$ , where a is the coefficient of $x^{2}$ and b is the coefficient of x. So we have,

$x=-\frac{b}{2a}=- \frac{400}{(2)(-40)}=5$

That means, the greatest revenue is achieved after 5 price increases. Each price increase was 20, so 5 price increase would be $5*20=100$ . So the price that gives the greatest revenue is $100+100=200$ .

ANSWERS:

A. $3960

B. $R=-40x^{2}+400x+3000$

C. $200

4 0

3 years ago

Please help me with this please

eimsori [14]

Part (a)

Using the quotient rule, the blank is 11/5.

Part (b)

Using the product rule, the blank is 9.

Part (c)

Using the power rule, the blank is 5.

6 0

1 year ago

What’s the slope of the line??

emmainna [20.7K]

If I did the math correctly I’d should be -5 or rewritten as -5/1.

8 0

3 years ago

Helpp pleaseeeeee eeeeee

Scilla [17]

Answer: The answer is D

Step-by-step explanation:

3 0

3 years ago

Im confused..I feel as if what I’m learning in class itself is not as difficult as the problems my teacher gives us to take home

RideAnS [48]

Ok, so remember that the derivitive of the position function is the velocty function and the derivitive of the velocity function is the accceleration function

x(t) is the positon function
so just take the derivitive of 3t/π +cos(t) twice
first derivitive is 3/π-sin(t)
2nd derivitive is -cos(t)
a(t)=-cos(t)

on the interval [π/2,5π/2) where does -cos(t)=1? or where does cos(t)=-1?
at t=π
so now plug that in for t in the position function to find the position at time t=π
x(π)=3(π)/π+cos(π)
x(π)=3-1
x(π)=2
so the position is 2

ok, that graph is the first derivitive of f(x)
the function f(x) is increaseing when the slope is positive
it is concave up when the 2nd derivitive of f(x) is positive

we are given f'(x), the derivitive of f(x)
we want to find where it is increasing AND where it is concave down

it is increasing when the derivitive is positive, so just find where the graph is positive (that's about from -2 to 4)

it is concave down when the second derivitive (aka derivitive of the first derivitive aka slope of the first derivitive) is negative
where is the slope negative?
from about x=0 to x=2
and that's in our range of being increasing
so the interval is (0,2)

4 0

3 years ago

P(x) Q(x) = R(x); if P(x) 72 Am R(x); if P(x)=x+2 and R(x) = x³ - 3x² - 6x-2, what is Q(x)?