The sum of two positive numbers is given. If | vedclass

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(A) Let $x$ and $y$ be two positive numbers. Given that their sum is constant,$x + y = a$. Let $z = x^3 + y^3$. Substituting $y = a - x$,we get $z = x

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they are equal

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(A) Let $x$ and $y$ be two positive numbers. Given that their sum is constant,$x + y = a$. Let $z = x^3 + y^3$. Substituting $y = a - x$,we get $z = x^3 + (a - x)^3$. Differentiating with respect to $x$: $\frac{dz}{dx} = 3x^2 - 3(a - x)^2 = 3(x^2 - (a^2 - 2ax + x^2)) = 3(2ax - a^2) = 3a(2x - a)$. Setting $\frac{dz}{dx} = 0$,we find $2x - a = 0$,which implies $x = \frac{a}{2}$. Since $y = a - x$,we get $y = a - \frac{a}{2} = \frac{a}{2}$. Checking the second derivative: $\frac{d^2z}{dx^2} = 6a$. Since $a > 0$,$\frac{d^2z}{dx^2} > 0$,confirming a minimum at $x = \frac{a}{2}$. Thus,$x = y$,meaning the numbers are equal.

The sum of two positive numbers is given. If the sum of their cubes is minimum,then

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