Question:

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

Updated On: Sep 8, 2023
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Solution and Explanation

The correct answer is Rs. 20.967.
Marginal cost is the rate of change of total cost with respect to output.
∴ Marginal cost (MC)\(=\frac{dc}{dx}=0.007(3x^2)-0.003(2x)+15\)
\(=0.021x^2-0.006x+15\)
When \(x = 17, MC = 0.021 (17^2 ) − 0.006 (17) + 15\)
\(= 0.021(289) − 0.006(17) + 15 \)
\(= 6.069 − 0.102 + 15 \)
\(= 20.967\)
Hence, when 17 units are produced, the marginal cost is Rs. 20.967.
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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

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