Find two positive numbers whose sum is 16 and the sum of whose cubes i

1.	Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Answer» Let one number be x. Then other number is (16 - x) Let S(x) = x³ +(16 - x)³ S¹ (x) = 3x² + 3(16 - x)² (-1) S¹¹(x) = 6x + 6(16 - x)¹(+1) S¹(x) = 0 gives, 3x² - 3 (16 - x)² =0 ⇒ x² - (16 - x)² = 0 ⇒ x² = (16 - x)² ⇒ x² - 256 + x² - 32x ⇒ 32x = 256 x = 256/32 x = 8 Also, S¹¹ (x) = 6(8) + 6(16 - 8) = 48 + 6(8) = 48 + 48 = 96 > 0 ∴ By second derivative test, x = 8 is the point of local minima of s. Hence sum of cubes of numbers is minimum when the numbers are 8 and 16 - 8 = 8

Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Answer»

Let one number be x. Then other number is (16 - x)

Let S(x) = x³ +(16 - x)³

S¹ (x) = 3x² + 3(16 - x)² (-1)

S¹¹(x) = 6x + 6(16 - x)¹(+1)

S¹(x) = 0 gives, 3x² - 3 (16 - x)² =0

⇒ x² - (16 - x)² = 0

⇒ x² = (16 - x)²

⇒ x² - 256 + x² - 32x

⇒ 32x = 256

x = 256/32

x = 8

Also, S¹¹ (x) = 6(8) + 6(16 - 8) = 48 + 6(8)

= 48 + 48 = 96 > 0

∴ By second derivative test, x = 8 is the point of local minima of s.

Hence sum of cubes of numbers is minimum when the numbers are 8 and 16 - 8 = 8

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Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

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