Question

The price p (in dollars) and the demand x for a particular steam iron are related by the equation x=1000-2p.

-What will the marginal revenue be when the revenue is maximized and why?
-Find the instantaneous rate of change of revenue at 400 steam irons.

Answer 1

Marginal revenue at max revenue = 0

Instantaneous rate of change of revenue at 400 steam irons is 100

Explanation:

We have the price-demand equation as

`x = 1000-2p`

Express p in terms of x

Adding 2p on both sides and subtracting x from both sides yields

`2p = 1000-x`

Dividing by 2 on both sides gives us

`p = 500-(x)/(2)`

The total revenue from selling x steam irons is given by

`R(x) = px = (500-(x)/(2))x = 500x - (x^2)/(2)` (1)

This revenue is maximized when the first differential is zero

First differential of
`R(x)` =
`R'(x) = (d)/(dx) (500x - (x^2)/(2)) = 500 - (2x)/(2) = 500 - x` (2)

Setting this equal to zero and solving for
`x` will give us the value of
`x` at which the revenue is maximum and therefore the marginal revenue

Setting
`500-x = 0`

gives

x = 500

At this demand level marginal revenue is 0

ie the number of steam irons to be sold to maximize revenue

The revenue at this demand level is given by plugging in this value into equation (1)

`R_(max) = 500(500) - (500^2)/(2) = 500^2 - (500^2)/(2) = (500^2)/(2) = 125,000`

At x = 400, the instantaneous rate of change is given by plugging in this value into Equation (2)

`R'(400) = 500-400 = 100`

and the revenue at x = 400 is obtained by plugging in 400 for the value of x in Eq 1

`R(400) = (500)(400) - (400^2)/(2) = 200,000 - 80,000 = 120,000`

Hope that helps