
The money to be spent for the we fare of the employees of a firm is proportional to the rate of change of its total revenue (Marginal revenue). If the total revenue (in rupees) received from the sale of \[x\] units of a product is given by $R(x) = 3{x^2} + 36x + 5$ , find the marginal revenue, when $x = 5$, and write value does the question indicate.
Answer
585k+ views
Hint:Here, we are going to take derivative with respect to x., Also using formula ${\frac{{d({x^n})}}{{dx}} = n{x^{n - 1}}}$
Complete step by step solution:
Given that: The total revenue (in rupees) received from the sale of units of a product is given by $R(x) = 3{x^2} + 36x + 5$
Here, Marginal Revenue = $\frac{{dR}}{{dx}}$
$\begin{array}{l}
R(x) = 3{x^2} + 36x + 5\\
\frac{{dR(x)}}{{dx}} = \frac{{d(3{x^2} + 36x + 5)}}{{dx}}
\end{array}$
$\begin{array}{l}
\frac{{dR(x)}}{{dx}} = 6x + 36\\
\\
\frac{{dR(x)}}{{dx}} = 6(5) + 36\\
\frac{{dR(x)}}{{dx}} = 30 + 36
\end{array}$ [ when $x = 5$, Marginal Revenue will be]
Marginal Revenue $ = 66$
Hence, the required marginal revenue is Rupees 66.
Additional Information: Always remember the five basic differentiation rule.
The constant rule
For Example: Derivative of any constant number is zero.
$\begin{array}{l}
f(x) = c\\
f'(x) = 0
\end{array}$
The power rule
For Example: power is multiplied in this form.
$\begin{array}{l}
f(x) = {x^n}\\
f'(x) = n.{x^{n - 1}}
\end{array}$
3.) The constant multiple rule
For Example: Always take constant aside before applying derivative to variables.
$\begin{array}{l}
f(x) = 5{x^{^3}}\\
f'(x) = 5.f'({x^3})\\
f'(x) = 5 \times 3 \times {x^2}\\
f'(x) = 15{x^2}
\end{array}$
4) The sum rule
For Example: derivative is applied in each term separately.
$\begin{array}{l}
f(x) = {x^4} + {x^3} + 2x\\
f'(x) = f'({x^4}) + f'({x^3}) + 2f'(x)
\end{array}$
5) The difference rule
For Example: Derivative is applied in each term separately.
$\begin{array}{l}
f(x) = 2{x^4} - 5{x^2} - 8x\\
f'(x) = 2f'({x^4}) - 5f'({x^2}) - 8f(x)
\end{array}$
Note: Always solve by taking consideration of the respective derivatives with the given substitution and simplified form would be the answer.
Complete step by step solution:
Given that: The total revenue (in rupees) received from the sale of units of a product is given by $R(x) = 3{x^2} + 36x + 5$
Here, Marginal Revenue = $\frac{{dR}}{{dx}}$
$\begin{array}{l}
R(x) = 3{x^2} + 36x + 5\\
\frac{{dR(x)}}{{dx}} = \frac{{d(3{x^2} + 36x + 5)}}{{dx}}
\end{array}$
$\begin{array}{l}
\frac{{dR(x)}}{{dx}} = 6x + 36\\
\\
\frac{{dR(x)}}{{dx}} = 6(5) + 36\\
\frac{{dR(x)}}{{dx}} = 30 + 36
\end{array}$ [ when $x = 5$, Marginal Revenue will be]
Marginal Revenue $ = 66$
Hence, the required marginal revenue is Rupees 66.
Additional Information: Always remember the five basic differentiation rule.
The constant rule
For Example: Derivative of any constant number is zero.
$\begin{array}{l}
f(x) = c\\
f'(x) = 0
\end{array}$
The power rule
For Example: power is multiplied in this form.
$\begin{array}{l}
f(x) = {x^n}\\
f'(x) = n.{x^{n - 1}}
\end{array}$
3.) The constant multiple rule
For Example: Always take constant aside before applying derivative to variables.
$\begin{array}{l}
f(x) = 5{x^{^3}}\\
f'(x) = 5.f'({x^3})\\
f'(x) = 5 \times 3 \times {x^2}\\
f'(x) = 15{x^2}
\end{array}$
4) The sum rule
For Example: derivative is applied in each term separately.
$\begin{array}{l}
f(x) = {x^4} + {x^3} + 2x\\
f'(x) = f'({x^4}) + f'({x^3}) + 2f'(x)
\end{array}$
5) The difference rule
For Example: Derivative is applied in each term separately.
$\begin{array}{l}
f(x) = 2{x^4} - 5{x^2} - 8x\\
f'(x) = 2f'({x^4}) - 5f'({x^2}) - 8f(x)
\end{array}$
Note: Always solve by taking consideration of the respective derivatives with the given substitution and simplified form would be the answer.
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