Question 36 (iii) (A) - CBSE Class 12 Sample Paper for 2025 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Feb. 13, 2025 by Teachoo

Question 36 (iii) (A) - For what value of, x volume of each container - CBSE Class 12 Sample Paper for 2025 Boards

part 2 - Question 36 (iii) (A) - CBSE Class 12 Sample Paper for 2025 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

part 3 - Question 36 (iii) (A) - CBSE Class 12 Sample Paper for 2025 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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Question 36 (iii) (A) For what value of 𝒙, the volume of each container is maximum?Now, V = 2(2𝑥^3−65𝑥^2+500𝑥) And, 𝐝𝐕/𝒅𝒙= 4(𝑥−5)(3𝑥−50) Putting 𝐝𝐕/𝒅𝒙= 0 4(𝑥−5)(3𝑥−50)=0 So, x = 5 and x = 𝟓𝟎/𝟑 If 𝒙 = 𝟓𝟎/𝟑 Breadth of box = 25 – 2𝑥 = 25 – 2(𝟓𝟎/𝟑) = 25 – 33.3 = –8.3 Since, breadth cannot be negative, ∴ x = 𝟓𝟎/𝟑 is not possible Hence, 𝒙 = 5 only Finding V’’(𝒙) V’(𝑥)=" 4" [𝟑𝒙^𝟐−𝟔𝟓𝒙+𝟐𝟓𝟎] V’’(𝑥)=4[6𝑥−65] Putting 𝒙=𝟓 V’’(𝟓)=4(6(5)−65)= 4(30−65)= 4(−35)= –140 V’’(𝒙)<𝟎 when 𝑥=5 Thus, V(𝑥) is maximum at 𝑥=5 ∴ Square of side 5 cm is cut off from each Corner

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 15 years. He provides courses for Maths, Science and Computer Science at Teachoo

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