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Find two positive numbers whose sum is 16 and thesum of whose cubes is minimum. |
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Answer» Let a number be x then other number is (16 - x). Let the sum of their cubes is S. Then ` S = x^(3) + (16 - x)^(3)` ` rArr (dS)/(dx) = 3x^(2)+3(16-x)^(2)(-1)` ` = 3x^(2) - 3(16-x)^(2)` and ` ((d^(2)S)/(dx^(2))) = 6x + 6(16-x) = 96` For minimum value ` (dS)/(dx) = 0` ` rArr 3x^(2) -3 (16-x)^(2)= 0` ` rArr x^(2) - (256+x^(2)-32x)=0` ` rArr 32x = 256 rArr x = 8` at ` x = 8, ((d^(2)S)/(dx^(2)))_(x=8) = 96 gt 0 ` ` :. ` at x = 8, S is minimum. `:. 16 - x = 16 - 8 = 8 ` Therefore numbers are 8 and 8. |
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