Skip to content
Worked Examples · Example 41

Q.A manufacturer produces x pants per week at total cost of ₹(x2+78x+2500)(x^2 + 78x + 2500). The price per unit is given by 8x=600−p8x = 600 - p, where 'p' is the price of each set. Find the maximum profit obtained, where the profit function is given by P(x)=R(x)−C(x)P(x) = R(x) - C(x).

Chandigarh CbseNCERTSubjective· 5mImportance★★★★★
47% · 41/87 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 →

With p=600−8x,p=600-8x, revenue R=600x−8x2R=600x-8x^2 and profit P(x)=−9x2+522x−2500,P(x)=-9x^2+522x-2500, maximised at x=29,x=29, giving P=₹5069.P=₹5069.

Profit P(x)=R(x)−C(x)P(x)=R(x)-C(x) where Revenue R(x)=p⋅xR(x)=p\cdot x (price ×\times quantity). Maximise with P′(x)=0,P'(x)=0, P′′(x)<0.P''(x)<0.

  1. Price from the demand relation: 8x=600−p⇒p=600−8x8x=600-p\Rightarrow p=600-8x (₹ per pant).
  2. Revenue: R(x)=p⋅x=(600−8x)x=600x−8x2.R(x)=p\cdot x=(600-8x)x=600x-8x^2.
  3. Cost (given): C(x)=x2+78x+2500.C(x)=x^2+78x+2500.
  4. Profit function: P(x)=R(x)−C(x)=600x−8x2−(x2+78x+2500)=−9x2+522x−2500.P(x)=R(x)-C(x)=600x-8x^2-(x^2+78x+2500)=-9x^2+522x-2500.
  5. Differentiate: P′(x)=−18x+522.P'(x)=-18x+522. …

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.