The minimum value of (Ax+By) , where xy=C^2 , for A, B, C > 0 is :

The correct answer is Option (3) → \( 2\sqrt{AB}\,C \)

The question intends x and y to be positive (as in the usual application to lengths and quantities); for negative x and y, Ax + By has no minimum.

Put \( y=\frac{C^2}{x} \), x > 0:

    \( \displaystyle S=Ax+\frac{BC^2}{x} \)

    \( \displaystyle \frac{dS}{dx}=A-\frac{BC^2}{x^2} \)

Setting \( \frac{dS}{dx}=0 \):

    \( \displaystyle x^2=\frac{BC^2}{A} \)

    \( \displaystyle x=C\sqrt{\frac{B}{A}} \)

    \( \displaystyle \frac{d^2S}{dx^2}=\frac{2BC^2}{x^3}\gt 0 \)

so S has its minimum there. At this x, \( y=\frac{C^2}{x}=C\sqrt{\frac{A}{B}} \), and

    \( \displaystyle S_{\min}=AC\sqrt{\frac{B}{A}}+BC\sqrt{\frac{A}{B}} \)

    \( \displaystyle =C\sqrt{AB}+C\sqrt{AB} \)

    \( \displaystyle =2\sqrt{AB}\,C \)

Shortcut: By AM ≥ GM for the positive numbers Ax and By:

    \( \displaystyle Ax+By\ge 2\sqrt{Ax\cdot By} \)

    \( \displaystyle =2\sqrt{AB\cdot xy}=2\sqrt{AB}\,C \)

with equality when Ax = By.