Deriving marginal cost function using first order derivative

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  First order derivative approach used to derive optimal quantity by minimizing marginal cost function

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  In our case it is given by: (1.26) And at the break even quantities: (1.27) (e) The marginal revenue function is the derivative of R(x) with respect to x: (1.28) (f) The marginal cost function is the derivative of C(x) with respect

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  In order to find it, we must find the first derivative of C with respect to L and equate it to 0.

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  The first thing to solve such questions is to write down all the equations: Q = 1500L + 60L² - L³ Taking the first order derivative we get the Marginal Product So MPl = 1500 + 120L - 3L² Also dividing Q by L we get the average product: APl = 1500

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  69702 First Derivative Test and Maximizing Revenue Functions Using the first derivative test, find the critical values and intervals where the function is increasing.

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  = dTC/dQ or MR = MC, because the Marginal Revenue MR is the first derivative of the Total Revenue and the Marginal Cost MC is the first derivative of Total Cost.

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  Question 1: As we know the function of the marginal cost is the first derivative of k.Let we use m(x) denote the marginal cost function.So m(x)=k`(x)=3x^2-18x+29.The function of m(x) will be minimized when quantity can make the first derivative of

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  b) Give a clear explanation using a difference quotient of why the given marginal cost equals the derivative C' (10,000), where C' (x) is the price in dollars of extracting x tons of ore. 5.