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  • Explain solution RD Sharma class 12 chapter Derivative as a Rate Measure exercise 12.1 question 8 maths

Explain solution RD Sharma class 12 chapter Derivative as a Rate Measure exercise 12.1 question 8 maths

Answers (1)

Answer: Rs 20.967

Hint:

To find marginal cost at given point, we have to differentiate the function with respect to its variable and put the value of variable in formula.

Given:

c(x)=0.007x^{3}-0.003x^{2}+15x+4000

Solution:

Here we have,

Marginal cost

c(x)=0.007x^{3}-0.003x^{2}+15x+4000

Let’s differentiate the c(x) with respect to x

\therefore \frac{dc(x)}{dx}=c'(x)

=3\times 0.007x^{2}-2\times 0.003x+15                \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]

Let's put x=17 units

\therefore c'(17)=3\times 0.007(17)^{2}-2\times 0.003\times 17+15

                =6.069-0.102+15

\therefore c'(17)=20.967 Rs

So 20.967 Rs is the marginal cost when 17 units are produced.

 

 

 

Posted by

Gurleen Kaur

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