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 August 14, 2026 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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