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

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(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

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$2c\sqrt{ab}$

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(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

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