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 Dec. 13, 2024 by Teachoo

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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