If xy = c^2,what is the minimum value of ax + | vedclass

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(C) Given $xy = c^2$,we have $y = \frac{c^2}{x}$.Let $f(x) = ax + by = ax + b\left(\frac{c^2}{x}\right) = ax + \frac{bc^2}{x}$.To find the minimum val

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$2c\sqrt{ab}$

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(C) Given $xy = c^2$,we have $y = \frac{c^2}{x}$.Let $f(x) = ax + by = ax + b\left(\frac{c^2}{x}\right) = ax + \frac{bc^2}{x}$.To find the minimum value,we differentiate $f(x)$ with respect to $x$:$f'(x) = a - \frac{bc^2}{x^2}$.Setting $f'(x) = 0$ for critical points:$a = \frac{bc^2}{x^2} \implies x^2 = \frac{bc^2}{a} \implies x = c\sqrt{\frac{b}{a}}$ (since $x, a, b, c > 0$ for a minimum in this context).Now,find the second derivative:$f''(x) = \frac{2bc^2}{x^3}$.Since $x > 0$,$f''(x) > 0$,confirming a minimum at $x = c\sqrt{\frac{b}{a}}$.Substituting $x$ back into $f(x)$:$f\left(c\sqrt{\frac{b}{a}}\right) = a\left(c\sqrt{\frac{b}{a}}\right) + \frac{bc^2}{c\sqrt{\frac{b}{a}}} = c\sqrt{ab} + c\sqrt{\frac{b^2}{b/a}} = c\sqrt{ab} + c\sqrt{ab} = 2c\sqrt{ab}$.

If $xy = c^2$,what is the minimum value of $ax + by$ (where $a > 0, b > 0$)?

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