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3.4 · Q6

Q.The price 'p' per unit is given by the relation x=13p2−2p+3x = \dfrac{1}{3}p^2 - 2p + 3 where 'x' is the number of units sold then, i. Find the revenue function R.
ii. Find the price interval for which the revenue is increasing and decreasing.

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
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Revenue is price ×\times quantity, giving R(p)=13p3−2p2+3pR(p)=\tfrac13 p^3-2p^2+3p; its derivative R′(p)=(p−1)(p−3)R'(p)=(p-1)(p-3) fixes the increasing/decreasing price ranges.

R=p⋅xR=p\cdot x, where pp = price per unit and xx = units sold. RR is increasing where R′(p)>0R'(p)>0 and decreasing where R′(p)<0R'(p)<0.

  1. Given x=13p2−2p+3x=\dfrac{1}{3}p^2-2p+3.
  2. Revenue R(p)=p⋅x=p(13p2−2p+3)=13p3−2p2+3pR(p)=p\cdot x=p\left(\dfrac{1}{3}p^2-2p+3\right)=\dfrac{1}{3}p^3-2p^2+3p.
  3. Differentiate: R′(p)=p2−4p+3=(p−1)(p−3)R'(p)=p^2-4p+3=(p-1)(p-3).
  4. Set R′(p)=0⇒p=1R'(p)=0\Rightarrow p=1 or p=3p=3 (the critical prices).
  5. Test the sign of R′(p)=(p−1)(p−3)R'(p)=(p-1)(p-3) on each interval: …

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