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The variable cost (V) of manufacturing a product varies according to the equation V= 4q, where q is the quantity produced. The fixed cost (F) of production of same product reduces with q according to the equation F = 100/q. How many units should be produced to minimize the total cost (V+F)?


1. 5
2. 4
3. 7
4. 6

1 Answer

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Best answer
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.

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