A company has the cost function $C (x) = 1000 + 50 x + 2 x^{2}$ rupees, where $x$ is the number of units produced in thousands. Determine the production level $x$ that minimises the average cost and compute the minimum average cost.

hard 4 marks Class 12 · Mathematics · Application of Derivatives

Answer

x = 10 s q r t 5

thousand units; minimum average cost

40 s q r t 5 + 50

rupees

The correct answer is $x = 10 s q r t 5$ thousand units; minimum average cost $40 s q r t 5 + 50$ rupees — Thus producing $10 s q r t 5 a pp r o x 22.36$ thousand units minimises the average cost, which equals $40 s q r t 5 + 50 a pp r o x 139.44$ rupees per unit.

Step-by-step solution

Average cost per unit: $A C (x) = \frac{C ( x )}{x} = \frac{1000}{x} + 50 + 2 x$ .
Differentiate: $A C^{'} (x) = - \frac{1000}{x ^{2}} + 2$ . Set to zero: $- \frac{1000}{x ^{2}} + 2 = 0; \Rightarrow; 2 = \frac{1000}{x ^{2}}; \Rightarrow; x^{2} = 500; \Rightarrow; x = 105$ (positive root).
Check that this gives a minimum by observing $A C^{''} (x) = \frac{2000}{x ^{3}} > 0$ .
Compute the minimum average cost: $A C (105) = \frac{1000}{10 5} + 50 + 2 (105) = 205 + 50 + 205 = 405 + 50$ rupees.
Thus producing $10 s q r t 5 a pp r o x 22.36$ thousand units minimises the average cost, which equals $40 s q r t 5 + 50 a pp r o x 139.44$ rupees per unit.

Common mistake: Treating total cost as average cost when differentiating

Concept tested: Optimising average cost

Published 2 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

A company has the cost function C(x)=1000+50x+2x2 rupees, where x is the number of units produced in thousands. Determine the production level x that minimises the average cost and compute the minimum average cost.

Step-by-step solution

Related questions

A company has the cost function $C (x) = 1000 + 50 x + 2 x^{2}$ rupees, where $x$ is the number of units produced in thousands. Determine the production level $x$ that minimises the average cost and compute the minimum average cost.