#### Provide solution for RD Sharma maths class 12 chapter Derivative as a Rate Measure exercise 12.1 question 10

$M\! R=Rs66$. It indicates the extra money spent when number of employees increase from 5 to 6

Given:

$R(x)=3x^{2}+36x+5$

Solution:

Here we have,

Total revenue

$R(x)=3x^{2}+36x+5$

To find marginal revenue $M\! R$ let’s differentiate R($x$) with respect to $x$

$\therefore \frac{d}{dx}R(x)=M\! R=6x+36$                                $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

To find $M\! R$

Put $x=5$ in $M\! R$

$\therefore M\! R=6(5)+36$

$\therefore M\! R=66\; Rs$

Here we can write

$M\! R=\lim _{h \rightarrow 1} \frac{R(x+h)-R(x)}{h}=66$

So, we can say thet in first there were 5 employees but after they become 6 employees.

Note:

Here we have taken

$M\! R=\lim _{h \rightarrow 1} \frac{R(x+h)-R(x)}{h}$

$N\! ot \; \lim _{h \rightarrow 0}\; because\; h\; represents$

Number of employees which will always be integer and greater than zero.