If xy = c^2,then the minimum value of ax + by | vedclass

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

Question

(B) Given $xy = c^2$,we can write $y = \frac{c^2}{x}$.Let $f(x) = ax + by = ax + \frac{bc^2}{x}$.To find the minimum value,differentiate $f(x)$ with r

Vedclass · Accepted Answer

$2c\sqrt{ab}$

Vedclass · Answer

(B) Given $xy = c^2$,we can write $y = \frac{c^2}{x}$.Let $f(x) = ax + by = ax + \frac{bc^2}{x}$.To find the minimum value,differentiate $f(x)$ with respect to $x$:$f'(x) = a - \frac{bc^2}{x^2}$.Set $f'(x) = 0$ for critical points:$a - \frac{bc^2}{x^2} = 0 \implies ax^2 = bc^2 \implies x^2 = \frac{bc^2}{a}$.Since $x$ must be positive for a minimum value in this context (assuming $a, b, c > 0$),we take $x = c\sqrt{\frac{b}{a}}$.Substituting $x$ back into $f(x)$:$f\left(c\sqrt{\frac{b}{a}}\right) = a\left(c\sqrt{\frac{b}{a}}\right) + \frac{bc^2}{c\sqrt{\frac{b}{a}}} = ac\sqrt{\frac{b}{a}} + bc\sqrt{\frac{a}{b}} = c\sqrt{ab} + c\sqrt{ab} = 2c\sqrt{ab}$.

If $xy = c^2$,then the minimum value of $ax + by$ is

Similar Questions

On the interval $[0,1]$,the function $f(x) = x^{25}(1-x)^{75}$ takes its maximum value at the point

Let $f(x) = 2 \cos^{-1} x + 4 \cot^{-1} x - 3x^2 - 2x + 10$,where $x \in [-1, 1]$. If $[a, b]$ is the range of the function,then $4a - b$ is equal to:

In the interval $[-2, 4]$,the absolute maximum of $f(x) = 2x^3 - 3x^2 - 12x + 5$ occurs at $x =$

If $x + y = 10$,then the maximum value of $xy$ is

For a positive real number $x$,when its reciprocal is added to it,the sum attains a minimum value. What is the value of $x$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module