The sum of two positive numbers is equal to $a$ and if the sum of their cubes is the least, the numbers are:

\frac{a}{3}, \frac{a}{3}

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\frac{a}{2}, \frac{a}{2}

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\frac{a}{4}, \frac{a}{4}

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a, \frac{a}{4}

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Solution

The correct option is B $\frac{a}{2}, \frac{a}{2}$
Let one of the number be 'x'.
Then another number will be $(a - x)$ .
Hence
$f (x) = x^{3} + (a - x)^{3}$
Differentiating with respect to x, gives us
$f^{'} (x) = 3 x^{2} - 3 (a - x)^{2}$
$f^{'} (x) = 0$ implies
$x^{2} - (a - x)^{2} = 0$
$(2 x - a) (a) = 0$
Since $a \neq 0$
Hence
$x = \frac{a}{2}$ .
Therefore both the numbers are
$\frac{a}{2}, \frac{a}{2}$ .

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The sum of two positive numbers is equal to a and if the sum of their cubes is the least, the numbers are:

The correct option is B a2,a2Let one of the number be 'x'.Then another number will be (a−x).Hence f(x)=x3+(a−x)3Differentiating with respect to x, gives us f′(x)=3x2−3(a−x)2f′(x)=0 implies x2−(a−x)2=0(2x−a)(a)=0Since a≠0Hence x=a2.Therefore both the numbers are a2,a2.

The sum of two positive numbers is equal to $a$ and if the sum of their cubes is the least, the numbers are: