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Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12 Important
Ex 6.3,13 You are here
Ex 6.3,14 Important
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
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Ex 6.3, 20 Important
Ex 6.3,21
Ex 6.3,22 Important
Ex 6.3,23 Important
Ex 6.3,24 Important
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Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Last updated at May 29, 2023 by Teachoo
Ex 6.3, 13 Find two numbers whose sum is 24 and whose product is as large as possible. Let first number be π₯ Now, given that First number + Second number = 24 π₯ + second number = 24 Second number = 24 β π₯ Product = (ππππ π‘ ππ’ππππ ) Γ (π πππππ ππ’ππππ) = π₯ (24βπ₯) Let P(π₯) = π₯ (24βπ₯) We need product as large as possible Hence we need to find maximum value of P(π₯) Finding Pβ(x) P(π₯)=π₯(24βπ₯) P(π₯)=24π₯βπ₯^2 Pβ(π₯)=24β2π₯ Pβ(π₯)=2(12βπ₯) Putting Pβ(π₯)=0 2(12βπ₯)=0 12 β π₯ = 0 π₯ = 12 Finding Pββ(π₯) Pβ(π₯)=24β2π₯ Pββ(π₯) = 0 β 2 = β 2 Thus, pββ(π₯) < 0 for π₯ = 12 π₯ = 12 is point of maxima & P(π₯) is maximum at π₯ = 12 Finding maximum P(x) P(π₯)=π₯(24βπ₯) Putting π₯ = 12 p(12)= 12(24β12) = 12(12) = 144 β΄ First number = x = 12 & Second number = 24 β x = (24 β 12)= 12