Question

Bea T. Howen, a sophomore college student, lost her scholarship after receiving a D in her "Music Appreciation" course. She decided to buy a snow plow to supplement her income during the winter months. It cost her $6312.50. Fuel and standard maintenance will cost her an additional $8.75 for each hour of use. Find the cost function C(x) associated with operating the snow plow for x hours. C(x)= If she charges $34.00 per hour write the revenue function R(x) for the amount of revenue gained from x hours of use. R(x)= Find the profit function P(x) for the amount of profit gained from x hours of use. P(x)= How many hours will she need to work to break even? hours

Answer 1

`(a)\ C(x) = 6312.50 + 8.75x`

`(b)\ R(x) = 34x`

`(c)\ P(x) = 6312.50 -25.25x`

`(d)\ Break\ Even = 250\ hours`

Explanation:

Solving (a): The cost function.

Given

`Cost = \$6312.50` --- cost of saxophone

`Additional = \$8.75` per hour

The cost function is: sum of the cost of the saxophone and the extra cost per hour.

If an hour costs 8.75, then x hours will cost 8.75x

So, the cost function is:

`C(x) = Cost + Additional`

`C(x) = 6312.50 + 8.75x`

Solving (b): The revenue function

Given

`Charges = \$34.00` per hour

The revenue function is the product of the unit charge by the number of hours.

If in an hour, she charges $34.00, then in x hours, she will cost 34x

So, the revenue function is:

`R(x) = Charges * Hours`

`R(x) = 34.00 * x`

`R(x) = 34x`

Solving (c): The profit function

This is the difference between the cost function and the revenue function

i.e.

`P(x) = C(x) - R(x)`

So, we have:

`P(x) = 6312.50 + 8.75x - 34x`

`P(x) = 6312.50 -25.25x`

Solving (d): The break even hours.

To do this, we simply equate the cost function and the revenue function, then solve for x

i.e.

`C(x) =R(x)`

`6312.50 + 8.75x = 34x`

Collect like terms

`6312.50 =- 8.75x + 34x`

`6312.50 = 25.25x`

Solve for x

`x = (6312.50)/(25.25)`

`x = 250`