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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 $F=\dfrac{100}{q}$. How many units should be produced to minimize the total cost $(V+F)$?

  1. $5$
  2. $4$
  3. $7$
  4. $6$
asked in Numerical Ability by Veteran (115k points)
edited by | 1.5k views

3 Answers

+19 votes
Best answer
Total Cost, $T = 4q +\dfrac{100}{q}$

When total cost becomes minimum, first derivative of $T$ becomes $0$ and second derivative at the minimum point will be positive.

Differentiating $T$ with respect to $q$ and equating to $0,$

$4 - \dfrac{100}{q^{2}} = 0\Rightarrow q = +5$  or $-5.$ Since, we can't have negative number of product, $q = 5.$

Taking second derivative, at $q = 5$ gives $\dfrac{200}{125} =\dfrac{8}{5} > 0,$ and hence $5$ is the minimum point.
answered by Veteran (384k points)
edited by
0

The function has a minimum value at x = a  if f '(a) = 0 
and f ''(a) = a positive number.

The function has a maximum value at x = a  if f '(a) = 0 
and f ''(a) = a negative number.

0 votes

is it A?

answered by (447 points) 1 flag:
✌ Edit necessary (syncronizing “that's not a comment section.”)
0 votes
One more approach is to substitute values of q as given options to check the min value.

if we take q=5 we get 40 in the equation of T=4q+(100/q) the other options give a value above 40.Just an approach handy during exams if we forget differentiation.
answered by Active (1.5k points)
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