Question

Part g: what price should the company charge to earn at least $4032 in revenue?

Answer 1

Given:

price p = $25

number of x units = -8p + 400

Find: solve for the value of x

Solution:

To determine the value of x, simply replace the variable "p" in the equation with 25.

`x=-8(25)+400`

Then, solve.

Multiply -8 and 25.

`x=-200+400`

Add -200 and 400.

`x=200`

Therefore, at $25, 200 units were sold.

Part F:

Since the equation of the revenue is R(x) = -8p² + 400p, then the graph must be a parabola opening down. Out of the 4 options, Only Options A and D show this.

However, upon comparing the two options, we see that the y-axis of the two graphs is different. Option A says that the y-axis is the price p while Option D says that the y-axis is the revenue R.

Since the given function is the revenue, then the y-axis should be R. The correct graph is Graph D.

Part G:

If the revenue is at least $4032, then let's replace the r(x) function with 4, 032.

`4,032=-8p^2+400p`

To solve for p, let's equate the function to zero by subtracting both sides of the function by 4, 032.

`\begin{gathered} 4,032-4032=-8p^2+400p-4032 \\ 0=-8p^2+400p-4032 \end{gathered}`

Then, let's solve for the value of "p" using the quadratic formula.

`p=(-b\pm√(b^2-4ac))/(2a)`

Note that in the revenue function above, a = -8, b = 400, and c = -4032. Let's plug these values into the formula above.

`p=(-400\pm√(400^2-4(-8)(-4032)))/(2(-8))`

Then, simplify.

`p=(-400\pm√(160,000-129,024))/(-16)`
`p=(-400\pm√(30,976))/(-16)`
`p=(-400\pm176)/(-16)`

Separate the plus and minus operations.

`p=(-400+176)/(-16)=(-224)/(-16)=14`
`p=(-400-176)/(-16)=(-576)/(-16)=36`

The values of p are 14 and 36.

Hence, the company should charge a price between a minimum of $14 and a maximum of $36 to have a revenue of at least $4,032.