Correct Answer - Option 1 : 5
Concept:
Total cost = Fixed cost + Variable cost
To find maxima and minima of a function y = f(x), follow these steps.
Step 1
\(Find\;\frac{{dy}}{{dx}},\;and\;put\frac{{dy}}{{dx}} = 0.\)
Find the value of x and this value is said to be the stationary point, this is a necessary condition to find the extremum value of a function.
Step 2
\(Find\;\frac{{{d^2}y}}{{d{x^2}}}\;\)
Check the value at the stationary point obtained in Step 1.
A function f(x) has a maxima at x = a if f’(a) = 0 and f”(a) < 0
A function f(x) has a minima at x = a if f’(a) = 0 and f”(a) > 0
A function f(x) has no maxima and minima at x = a if f’(a) = 0 and f”(a) = 0.
Calculation:
Given:
F = 100/q, V = 4q
Total cost (TC) = Fixed cost + Variable cost
\(∴\;TC=4q\;+\;\frac{100}{q}\)
Step 1:
\(\frac{{d(TC)}}{{dq}} = 0\)
\(\frac{d(TC)}{dq}=4\;-\;\frac{100}{q^2}\)
∴ q = ± 5
\(\frac{{d^2(TC)}}{{dq^2}} =+ve\;for\;minima\)
\(\frac{{d^2(TC)}}{{dq^2}} =-ve\;for\;maxima\)
\(\frac{d^2(TC)}{dq^2}=\frac{200}{q^3}\)
At q = 5
\(\frac{d^2(TC)}{dq^2}=\frac{200}{5^3}\Rightarrow1.6\;\;(+ve\;,\;∴\;minima)\)
At q = -5
\(\frac{d^2(TC)}{dq^2}=\frac{200}{5^3}\Rightarrow-1.6\;\;(-ve\;,\;∴\;maxima)\)
∴ at q = 5, Total cost (TC) will be minimum.