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

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

Question

(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

Vedclass · Accepted Answer

they are equal

Vedclass · Answer

(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

Similar Questions

Consider a square of side $2 \ cm$. It is cut from one of its corners as shown in the adjacent figure. The maximum value of the sum of the perimeters of the two plane figures thus formed is

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.

Find the maximum and minimum values of the function $f(x)=(2x-1)^{2}+3$.

If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha \log |x|+\beta x^2+x$,then

The shortest distance between the point $\left( \frac{3}{2}, 0 \right)$ and the curve $y = \sqrt{x}, (x > 0)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module