Skip to content
3.4 · Q8

Q.The price 'p' per unit at which a company can sell all that it produces is given by p = 29 – x, where 'x' is the number of units produced. The total cost function C(x) = 45 + 11x. If P(x) = R(x) – C(x), is the profit function then find the interval in which the profit is increasing and decreasing.

Ladakh CbseNCERTSubjective· 3mImportance★★★★★
94% · 82/87 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Form profit P=R−C=−x2+18x−45P=R-C=-x^2+18x-45; its derivative P′(x)=−2x+18P'(x)=-2x+18 is positive below x=9x=9 and negative above, giving the increasing/decreasing intervals.

Revenue R(x)=p⋅xR(x)=p\cdot x; Profit P(x)=R(x)−C(x)P(x)=R(x)-C(x). PP increases where P′(x)>0P'(x)>0 and decreases where P′(x)<0P'(x)<0.

  1. Given p=29−xp=29-x and C(x)=45+11xC(x)=45+11x.
  2. Revenue R(x)=p⋅x=(29−x)x=29x−x2R(x)=p\cdot x=(29-x)x=29x-x^2.
  3. Profit P(x)=R(x)−C(x)=(29x−x2)−(45+11x)=−x2+18x−45P(x)=R(x)-C(x)=(29x-x^2)-(45+11x)=-x^2+18x-45.
  4. Differentiate: P′(x)=−2x+18P'(x)=-2x+18.
  5. Set P′(x)=0⇒−2x+18=0⇒x=9P'(x)=0\Rightarrow -2x+18=0\Rightarrow x=9. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.