Help with homework please!

Question

1. 

The price p​ (in dollars) and the quantity x sold of a certain product obey the demand equation

P = -1/6x + 100.

Answer parts ​(a) through ​(e).

​(a) Find a model that expresses the revenue R as a function of x.​ (Remember, R = xp)

​R(x)= _ (use integers or fractions for any numbers in expression)

​(c) What is the revenue if 250 units are​ sold?

R≈​$ _ (round to nearest hundredth if needed)

​(d) What quantity x maximizes​ revenue?

x= _ units

What is the maximum​ revenue?

R=​$ _

​(e) What price should the company charge to maximize revenue?

p=​$ _

2.

The price p and the quantity x sold of a certain product obey the demand equation below.

x= -8p + 192, 0 ≤ p ≤ 24

(a) Express the revenue R as a function of x.

(b) What is the revenue if 176 units are sold?

(c) What quantity x maximizes revenue? What is the maximum revenue?

(d) What price should the company charge to maximize revenue?

(e) What price should the company charge to earn at least $952 in revenue?

(a) R(x) = _

(b) The revenue if 176 units are sold is $ _

(c)The quantity _ units maximizes revenue.

The maximum revenue is $ _

(d) The company should charge $ _ to maximize revenue.

(e) The company should charge at least $ _ to at most $ _ to earn at least $952 in revenue.

Answer 1

For the first problem revenue equal xp so

a) r(x)= x(-1/6x + 100)

= -1/6 x^2 + 100x

c) if x = 250 then plug 250 in for x in the revenue equation and evaluate

I got $14583.33

d) To find the max revenue, find the derivative of the revenue function, set it equal to zero and solve for x

r'(x) = -1/3 x + 100

-1/3 x + 100 = 0 when x = 33

You have to sell 33 items to maximize revenue

R(33) tells you what the max revenue is

e) p(33) tells you what price you need to charge p(33) = -1/6(33) + 100