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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.

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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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