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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?$

Sagot :

Solution

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

[tex]R(x)\text{ = 290x - 0.52x}^2[/tex]

Step 3

Find the first derivative

[tex]\begin{gathered} R(x)=\text{ 290x -0.52x}^2 \\ R^{\prime}(x)\text{ = 290 - 1.04x} \end{gathered}[/tex]

Step 4:

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

[tex]\begin{gathered} 290\text{ - 1.04x = 0} \\ 1.04x\text{ = 290} \\ \text{x = }\frac{290}{1.04} \\ \text{x = 278.8 }\approx\text{ 279} \end{gathered}[/tex]

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

Step 5:

Substituting this value in the original equation gives the revenue:

[tex]\begin{gathered} R\text{ = 290x - 0.52x}^2 \\ R\text{ = 290}\times279\text{ - 0.52 }\times\text{ 279}^2 \\ R\text{ = 80910 - 42034.14} \\ R\text{ = \$38875.86} \end{gathered}[/tex]

Maximum revenue = $38875.86