NCERT Class 12 · Ch 6 Exercise 6.3 · Q26
NCERT Class 12 Maths · Chapter 6 Applications Of Derivatives · Exercise 6.3 Question 26
Show that a cone of fixed total surface area and maximum volume has semi-vertical angle sin⁻¹(1/3).
💡 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: α=sin⁻¹(1/3).
- Put u=sin α=r/l, 0<u<1. For fixed total surface S=πr(l+r), r²=Su/[π(1+u)].
- V²=(S³/(9π))·u(1−u)/(1+u)². Maximizing V is equivalent to maximizing this positive square.
- The derivative of u(1−u)/(1+u)² is (1−3u)/(1+u)³, positive before u=1/3 and negative after it.
- The endpoints give zero limiting volume, so the maximum is at sin α=1/3.
Master Chapter 6: Applications Of Derivatives
Practise interactive questions, flashcards, visual mind maps, and timed board tests for Applications Of Derivatives.
← Previous Question
Show that a cone with fixed slant height and maximum volume has…
Next Question →Which listed point on x²=2y is nearest (0,5)?
All Questions in Applications Of Derivatives — Exercise 6.3
Related Questions in Applications Of Derivatives
Class 12 Applications Of Derivatives · Mcq Single Correct
Find the maximal open interval on which f(x)=x²e^(−x) has positive derivative.
Class 12 Applications Of Derivatives · Mcq Single Correct
For f(x)=x²−4x+6, identify the maximal open interval where f′(x)>0.
Class 12 Applications Of Derivatives · Mcq Single Correct
For x>0, at which x is f(x)=x^x stationary?
Class 12 Applications Of Derivatives · Mcq Single Correct
Find the maximum value of (1/x)^x for x>0.
Class 12 Applications Of Derivatives · Mcq Single Correct
Find the maximum of [x(x−1)+1]^(1/3) on 0≤x≤1.
Class 12 Applications Of Derivatives · Mcq Single Correct
A trapezium area model is A(x)=(10+x)√(100−x²) cm² for 0≤x≤10. Find its maximum.