Ex 6.3

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 š„ Now, given that First number + Second number = 24 š„ + second number = 24 Second number = 24 ā š„ Product = (šššš š” šš¢šššš ) Ć (š ššššš šš¢šššš) = š„ (24āš„) Let P(š„) = š„ (24āš„) We need product as large as possible Hence we need to find maximum value of P(š„) Finding Pā(x) P(š„)=š„(24āš„) P(š„)=24š„āš„^2 Pā(š„)=24ā2š„ Pā(š„)=2(12āš„) Putting Pā(š„)=0 2(12āš„)=0 12 ā š„ = 0 š„ = 12 Finding Pāā(š„) Pā(š„)=24ā2š„ Pāā(š„) = 0 ā 2 = ā 2 Thus, pāā(š„) < 0 for š„ = 12 š„ = 12 is point of maxima & P(š„) is maximum at š„ = 12 Finding maximum P(x) P(š„)=š„(24āš„) Putting š„ = 12 p(12)= 12(24ā12) = 12(12) = 144 ā“ First number = x = 12 & Second number = 24 ā x = (24 ā 12)= 12