BPSC 47th Exam Prelims 2005 Solved Paper

Question : 131 of 150

Marks: +1, -0

The demand function of a commodity is x

= 1/ 3 (24 – 2p)

, where x is the number of units demanded and p is per unit price. The value of p, which maximizes the revenue, is –

Solution:

From the question, Demand function

x=\;{1}/{3}(24-2 p)

⇒ 3 x=24-2 p

⇒ p=\;{24-3 x}/{2}

p=12-\;{3 x}/{2}

{R}=p(x)=12 x-\;{3 x^{2}}/{2}

For maximum value,

\;{d R}/{d x}=12-\;{3}/{2} × 2 x=12-3 x

\;{d R}/{d x}=0

12-3 x=0 ⇒ x=4

\;{d^{2} R}/{d x^{2}}=0-3=-3

So, at

x=4 {R}(x)

will be maximum

P=12-\;{3 × 4}/{2}=6

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