Application of Derivatives — Class 12 Maths Solution

ncert exercise SA NCERT Ex.6.1,Q.No. 15,Page 198

Question

The total cost $C(x)$ in rupees associated with the production of $x$ units of an item is given by $C(x) = 0.007{x^3} - 0.003{x^2} + 15x + 4000$.Find the marginal cost when $17$ units are produced.

Step-by-step Solution

We have, $C(x) = 0.007{x^3} - 0.003{x^2} + 15x + 4000$ …(i)

Differentiating (i) w.r.t. $x$, we get

Marginal cost $= \cfrac{{dC}}{{dx}} = 0.007 \times 3{x^2} - 0.003 \times 2x + 15 + 0$

Therefore the marginal cost when 17 units are produced $= Rs20.967$.

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NCERT & Exemplar solution for CBSE Class 12 Mathematics, Application of Derivatives. Curated by Sachin Sharma. Free for all students.