GATE CSE 2011 | Question: 63

Question

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 $(V+F)$?

• A. $5$
• B. $4$
• C. $7$
• D. $6$

Answer 1

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.

Correct Answer: $A$

Comments

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.

Answer 2

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.

Answer 3

Variable Cost (V)
             V = 4q
                │
                │
                │
        Fixed Cost (F)
             F = 100/q
                │
                │
                ▼
      Total Cost (V + F)
     TC = 4q + 100/q
                │
                │  Differentiate w.r.t q
                │
                ▼
        d(TC)/dq = 4 − 100/q²
                │
                │  For minimum cost:
                │  d(TC)/dq = 0
                ▼
           4 = 100/q²
                │
                │
                ▼
              q² = 25
                │
                ▼
        Minimum at q = 5 units