(A) The marginal cost function $MC$ is the derivative of the cost function $C(x)$ with respect to $x$.$MC = \frac{dC}{dx} = \frac{d}{dx}(0.05x^3 - 0.2

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(A) The marginal cost function $MC$ is the derivative of the cost function $C(x)$ with respect to $x$.$MC = \frac{dC}{dx} = \frac{d}{dx}(0.05x^3 - 0.2x^2 + 3x + 500)$.Applying the power rule, we get $MC = 0.15x^2 - 0.4x + 3$.To find the marginal cost at $x = 3$, we substitute $x = 3$ into the $MC$ function:$MC(3) = 0.15(3)^2 - 0.4(3) + 3$.$MC(3) = 0.15(9) - 1.2 + 3$.$MC(3) = 1.35 - 1.2 + 3$.$MC(3) = 0.15 + 3 = 3.15$.Thus, the marginal cost at $x = 3$ is $3.15$ Rupees.

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

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The total cost $C(x)$ in Rupees, associated with the production of $x$ units of an item is given by $C(x) = 0.05x^3 - 0.2x^2 + 3x + 500$. The marginal cost, where $x = 3$ is $\rule{1cm}{0.15mm}$ (in Rupees).

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A
$3.15$
B
$30.15$
C
$3.015$
D
$30.015$

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The total cost $C(x)$ in Rupees, associated with the production of $x$ units of an item is given by $C(x) = 0.05x^3 - 0.2x^2 + 3x + 500$. The marginal cost, where $x = 3$ is $\rule{1cm}{0.15mm}$ (in Rupees).

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$x$ and $y$ are the sides of two squares such that $y = x - x^{2}$. Find the rate of change of the area of the second square with respect to the area of the first square.

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