The number 28 is divided into two positive pa | vedclass

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(D) Let the two parts be $x$ and $28-x$, where $x > 0$ and $28-x > 0$. Let $f(x) = x^3 + (28-x)^2$. To find the minimum, differentiate $f(x)$ with res

Vedclass · Accepted Answer

$20$

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(D) Let the two parts be $x$ and $28-x$, where $x > 0$ and $28-x > 0$. Let $f(x) = x^3 + (28-x)^2$. To find the minimum, differentiate $f(x)$ with respect to $x$: $f'(x) = 3x^2 - 2(28-x) = 3x^2 + 2x - 56$. Set $f'(x) = 0$: $3x^2 + 2x - 56 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-2 \pm \sqrt{4 - 4(3)(-56)}}{6} = \frac{-2 \pm \sqrt{4 + 672}}{6} = \frac{-2 \pm \sqrt{676}}{6} = \frac{-2 \pm 26}{6}$. Since $x > 0$, $x = \frac{24}{6} = 4$. The two parts are $4$ and $28-4 = 24$. The absolute difference is $|24 - 4| = 20$.

The number $28$ is divided into two positive parts such that the sum of the cube of one part and the square of the other part is minimum. Find the absolute difference between the two parts.

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