NCERT Class 12 · Ch 6 Exercise 6.3 · Q6
NCERT Class 12 Maths · Chapter 6 Applications Of Derivatives · Exercise 6.3 Question 6
Find the maximum profit for p(x)=41−72x−18x².
💡 Concept Hints
- Identify the domain and the inner and outer functions, or the relevant one-sided limits.
- Apply the appropriate differentiation or limit rule, keep all chain factors, and check exceptional points.
✅ Step-by-Step Solution
Final Answer: On R: maximum 113 at x=−2. If production requires x≥0: maximum 41 at x=0.
- p′=−72−36x, so the unrestricted stationary point is x=−2.
- p″=−36<0 and p=113−18(x+2)², proving the unrestricted maximum is 113.
- If x means nonnegative output, the vertex is infeasible. On [0,∞), p′<0 and the maximum is p(0)=41.
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