Ex 6.3,13 - Chapter 6 Class 12 Application of Derivatives
Last updated at August 25, 2026 by Teachoo
Ex 6.3
Ex 6.3, 1 (i)
Important
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Chapter 6 Class 12 Application of Derivatives
Serial order wise
Transcript
Ex 6.3, 13 Find two numbers whose sum is 24 and whose product is as large as possible. Let first number be 𝑥 Now, given that First number + Second number = 24 𝑥 + second number = 24 Second number = 24 – 𝑥 Product = (𝑓𝑖𝑟𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 ) × (𝑠𝑒𝑐𝑜𝑛𝑑 𝑛𝑢𝑚𝑏𝑒𝑟) = 𝑥 (24−𝑥) Let P(𝑥) = 𝑥 (24−𝑥) We need product as large as possible Hence we need to find maximum value of P(𝑥) Finding P’(x) P(𝑥)=𝑥(24−𝑥) P(𝑥)=24𝑥−𝑥^2 P’(𝑥)=24−2𝑥 P’(𝑥)=2(12−𝑥) Putting P’(𝑥)=0 2(12−𝑥)=0 12 – 𝑥 = 0 𝑥 = 12 Finding P’’(𝑥) P’(𝑥)=24−2𝑥 P’’(𝑥) = 0 – 2 = – 2 Thus, p’’(𝑥) < 0 for 𝑥 = 12 𝑥 = 12 is point of maxima & P(𝑥) is maximum at 𝑥 = 12 Finding maximum P(x) P(𝑥)=𝑥(24−𝑥) Putting 𝑥 = 12 p(12)= 12(24−12) = 12(12) = 144 ∴ First number = x = 12 & Second number = 24 – x = (24 – 12)= 12
