(A) The marginal cost $(MC)$ is the derivative of the total cost function $C(x)$ with respect to $x$: $MC = \frac{dC}{dx} = \frac{d}{dx}(0.005x^{3} -

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(A) The marginal cost $(MC)$ is the derivative of the total cost function $C(x)$ with respect to $x$: $MC = \frac{dC}{dx} = \frac{d}{dx}(0.005x^{3} - 0.02x^{2} + 30x + 5000)$ $MC = 0.005(3x^{2}) - 0.02(2x) + 30$ $MC = 0.015x^{2} - 0.04x + 30$ To find the marginal cost when $x = 3$ units are produced,substitute $x = 3$ into the expression for $MC$: $MC = 0.015(3)^{2} - 0.04(3) + 30$ $MC = 0.015(9) - 0.12 + 30$ $MC = 0.135 - 0.12 + 30$ $MC = 0.015 + 30 = 30.015$ Thus,the marginal cost is $Rs. 30.015$.

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

Medium

The total cost $C(x)$ in Rupees,associated with the production of $x$ units of an item is given by $C(x) = 0.005x^{3} - 0.02x^{2} + 30x + 5000$. Find the marginal cost when $3$ units are produced,where marginal cost is defined as the instantaneous rate of change of total cost with respect to the number of units produced.

A
$Rs. 30.015$
B
$Rs. 30.02$
C
$Rs. 30.05$
D
$Rs. 30.10$

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Applications of Derivatives

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Rate of Change of Quantities

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