Max/min problem - show that diagonal is least when rectangle is a square. A rectangle has constant area, show that the length of a diagonal is least when the rectangle is a square.




$$Area (a) = xy$$

$$y=\dfrac{a}{x}$$

$$D^2 = x^2 + y^2$$

$$D^2 = x^2 + (\dfrac{a}{x})^2$$

$$\dfrac{dD^2}{dx} = 2x - \dfrac{2a^2}{x^3}$$

$$2x - (\dfrac{2a^2}{x^3})=0$$

$$x=a^\dfrac{2}{4}$$

$$x=\sqrt{a}$$

$$\dfrac{d^2D^2}{dx^2}=2+\dfrac{6a^2}{x^4}$$

$$2+\dfrac{6a^2}{x^4}&gt;0, minima$$

$$y=\dfrac{a}{\sqrt{a}}$$

$$y=\sqrt{a}$$

Is this correct? I'm assuming it is since the length of $x$ and $y$ are the same..sorry, I don't usually get these questions right and just wanted to make sure this is the way to do it.

Thank you.
the proof above is ok, since $x^2+y^2\geq 2xy$ and this is equivalent to $(x-y)^2\geq 0$ and the equal sign holds if $x=y$ this means we have a square.
$D =\sqrt{x^2+y^2} \geq \sqrt{2xy}=K$, constant. Thus equality occurs when $x = y$, or you have a square.