Question

The revenue function R in terms of the number of units sold, x, is given as R = 290x-0.52x^2where R is the total revenue in dollars. Find the number of units sold x that produces a maximum revenue?Your answer is x=What is the maximum revenue?$

Answer 1

Step 1:

The function reaches a maximum where the derivative is equal to 0.

Find the first derivative of the function.

Step 2:

Write the function

`R(x)\text{ = 290x - 0.52x}^2`

Step 3

Find the first derivative

`\begin{gathered} R(x)=\text{ 290x -0.52x}^2 \\ R^(\prime)(x)\text{ = 290 - 1.04x} \end{gathered}`

Step 4:

The function reaches a maximum where the derivative is equal to 0.

`\begin{gathered} 290\text{ - 1.04x = 0} \\ 1.04x\text{ = 290} \\ \text{x = }(290)/(1.04) \\ \text{x = 278.8 }\approx\text{ 279} \end{gathered}`

So the number of units which produce the maximum revenue = 279

Step 5:

Substituting this value in the original equation gives the revenue:

`\begin{gathered} R\text{ = 290x - 0.52x}^2 \\ R\text{ = 290}*279\text{ - 0.52 }*\text{ 279}^2 \\ R\text{ = 80910 - 42034.14} \\ R\text{ = \$38875.86} \end{gathered}`

Maximum revenue = $38875.86