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.
Find the marginal cost when 17 units are produced.
Solution and Explanation
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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Notes on Applications of Derivatives
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.
Read More: Application of Derivatives