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Question: The money to be spent for the we fare of the employees of a firm is proportional to the rate of chan...

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 xx units of a product is given by R(x)=3x2+36x+5R(x) = 3{x^2} + 36x + 5 , find the marginal revenue, when x=5x = 5, and write value does the question indicate.

Explanation

Solution

Here, we are going to take derivative with respect to x., Also using formula d(xn)dx=nxn1{\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)=3x2+36x+5R(x) = 3{x^2} + 36x + 5
Here, Marginal Revenue = dRdx\frac{{dR}}{{dx}}
R(x)=3x2+36x+5 dR(x)dx=d(3x2+36x+5)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}
dR(x)dx=6x+36  dR(x)dx=6(5)+36 dR(x)dx=30+36\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=5x = 5, Marginal Revenue will be]
Marginal Revenue =66 = 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.
f(x)=c f(x)=0\begin{array}{l} f(x) = c\\\ f'(x) = 0 \end{array}
The power rule
For Example: power is multiplied in this form.

f(x)=xn f(x)=n.xn1\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.
f(x)=5x3 f(x)=5.f(x3) f(x)=5×3×x2 f(x)=15x2\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.
f(x)=x4+x3+2x f(x)=f(x4)+f(x3)+2f(x)\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.
f(x)=2x45x28x f(x)=2f(x4)5f(x2)8f(x)\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.