Question

the demand equation for a certain product is given by p=116-0.015x where p is the unit price(in dollars) of the product and x is the number of units produced the total revenue obtained by producing and selling x units is give by R=xpDetermine price P that would yield a revenue of 7270 dollarslowest such price=_____dollarsHighest such price=____ dollars

Answer 1

In this case, we'll have to carry out several steps to find the solution.

Step 01:

Data

demand equation ==> p = 116 - 0.015 x

p= unit price ($)

x = number of units

total revenue ===> R = x p

R = $7270

Step 02:

R = x * p

7270 = x * p

7270 = x * ( 116 - 0.015 x)

7270 = 116 x - 0.015 x²

0.015 x² - 116 x + 7270 = 0

Step 03:

We must solve the roots

0.015 x² - 116 x + 7270 = 0

`x\text{ =}\frac{\text{ -b }\pm\text{ }\sqrt[]{b^2\text{ -4a}\cdot c}}{2.a}`

0.015 x² - 116 x + 7270 = 0

a = 0.015

b = -116

c = 7270

`x\text{ =}\frac{\text{ - (-116) }\pm\sqrt[]{(-116)^2-4\cdot0.015\cdot7270}}{2\cdot0.015}`
`x\text{ =}\frac{\text{ 116 }\pm114.104}{0.03}`

x 1 = (116 + 114.104) / 0.03 = 7670.13

x2 = (116 - 114.104) / 0.03 = 63.2

Step 04:

p1 = 116 - 0.015 * x1 = 116 - 0.015 * (7670.13) = 0.948

p2 = 116 - 0.015 * x2 = 116 - 0.015 * (63.2) = 115.052

The answer is:

lowest such price = $0.948

highest such price = $115.052