The number 28 is divided into two positive pa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$20$

Vedclass · Answer

(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.

Similar Questions

${a_1, a_2, ....., a_n, .....}$ is a progression where $a_n = \frac{n^2}{n^3 + 200}$. The largest term of this progression is

If a cylindrical vessel of given volume $V$ with no lid on the top is to be made from a sheet of metal,then the radius $(r)$ and height $(h)$ of the vessel so that the metal sheet used is minimum,is

If $4+3x-7x^2$ attains its maximum value $M$ at $x=\alpha$ and $5x^2-2x+1$ attains its minimum value $m$ at $x=\beta$, then $\frac{28(M-\alpha)}{5(m+\beta)}=$

If $x, y$ are two positive integers such that $x+y=20$ and the maximum value of $x^3 y$ is $k$ at $x=\alpha, y=\beta$,then $\frac{k}{\alpha^2 \beta^2} =$

If $x + y = 16$ and $x^2 + y^2$ is minimum,then the values of $x$ and $y$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module