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Last updated at April 19, 2021 by Teachoo

Example 34 Find two positive numbers whose sum is 15 and the sum of whose squares is minimum. Let first number be π Since Sum of two positive numbers is 15 π₯+ 2nd number = 15 2nd number = 15 β π Let S(π₯) be the sum of the squares of the numbers S(π₯)= (1st number)2 + (2nd number) 2 S(π)=π^π+(ππβπ)^π We need to minimize S(π) Finding Sβ(π) Sβ(π₯)=π(π₯^2+ (15 β π₯)^2 )/ππ₯ =π(π₯^2 )/ππ₯+(π(15 β π₯)^2)/ππ₯ = 2π₯+ 2(15βπ₯)(β1) = 2π₯β 2(15βπ₯) = 2π₯β30+2π₯ = 4πβππ Putting Sβ(π)=π 4π₯β30=0 4π₯=30 π₯=30/4 π=ππ/π Finding Sββ(π) Sββ(π₯)=π(4π₯ β 30)/ππ₯ = 4 Since Sββ(π)>π at π₯=15/2 β΄ π₯=15/2 is local minima Thus, S(π₯) is Minimum at π₯=15/2 Hence, 1st number = π₯=ππ/π 2nd number = 15βπ₯=15β15/2=ππ/π