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. Maximizing Revenue

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

x = -5p +100  0<p < or equal to 20
(a) Express the revenue R as a function of x.
(b) What is the revenue if 15 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
$480 in revenue?
asked Jan 19, 2015 in PRECALCULUS by anonymous

1 Answer

0 votes

Step 1 :

Here,

is the price in dollars.

is the quantity sold of a certain product .

The demand equation is .

Solve for .

.

Step 2 :

(a)

The revenue .

Substitute in .

Revenue .

Step 3 :

(b)

The quantity sold of a certain product .

Revenue .

The revenue if 15 units are sold is.

Step 4 :

(c)

The function is a quadratic function.

Compare the function with standard form of a quadratic function.

.

Since , the vertex is the highest point on the parabola.

The revenue is a maximum when the quantity sold of a certain product is .

image

Maximum revenue,

image

Maximum revenue isimage

answered Jan 20, 2015 by lilly Expert

Contd....

Step 5 :

(d)

The price .

Maximum revenue isimage at image.

At image, the company charge to maximum price.

The maximum price,

image

image should the company charge to maximize the revenue.

Step 6 :

(e)

Graph and are on the same Cartesian plane.

Find where the graphs intersects.

The graph is shown below :

 image

The graphs intersect at .

From the graph the company should charge between to earn at least in revenue.

Solution:

(a) Revenue .

(b) The revenue if 15 units are sold is.

(c) The maximum quantity is image and the maximum revenue isimage

(d) image should the company charge to maximize the revenue.

(e) The company should charge between to earn at least in revenue.

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