Question

The total revenue function for a product is given by R=520x dollars, and the total cost function for this same product is given by C=16,000+80x+x2, where C is measured in dollars. For both functions, the input x is the number of units produced and sold.a. Form the profit function for this product from the two given functions.b. What is the profit when 30 units are produced and sold?c. What is the profit when 45 units are produced and sold?d. How many units must be sold to break even on this product?

Answer 1

(a)

The profit function, P(x), is

Profit = Revenue - Cost

So,

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

We can find the simplified profit function, shown below:

`\begin{gathered} P(x)=R(x)-C(x) \\ P(x)=520x-(16000+80x+x^2) \\ P(x)=520x-16000-80x-x^2 \\ P(x)=-x^2+440x-16000 \end{gathered}`

(b)

To find the profit when 30 units are sold, we substitute x = 30 into the profit function.

So,

`\begin{gathered} P(x)=-x^2+440x-16000 \\ P(30)=-(30)^2+440(30)-16000 \\ P(30)=-3700 \end{gathered}`

So, there is a loss of $3700.

Or, profit of $ -3700

(c)

To find the profit when 45 units are sold, we substitute x = 45 into the profit function.

So,

`\begin{gathered} P(x)=-x^2+440x-16000 \\ P(45)=-(45)^2+440(45)-16000 \\ P(45)=1775 \end{gathered}`

The profit is $ 1775

(d)

Break Even is the point where Revenue equal Cost.

So,

`R(x)=C(x)`

Let's find the number of units to break-even,

`\begin{gathered} R(x)=C(x) \\ 520x=16,000+80x+x^2 \\ 16,000+80x+x^2-520x=0 \\ x^2-440x+16,000=0 \\ (x-40)(x-400)=0 \\ x=40,400 \end{gathered}`

For x = 40 units, it will break even.