# Find the rectangle with the maximum area, which can be turned in the corner. ?

## Find the rectangle with the maximum area, which can be turned in the corner

Jun 28, 2018

The rectangle will have a length of $\sqrt{2}$ , a width of $\frac{\sqrt{2}}{2}$ and thus area of $1 {m}^{2}$ #### Explanation:

From the sketch it is seen that the length AB = BC = $\sqrt{2}$ [ by Pythagoras] since the 'tightest' position the rectangle can be in is when it's base is at 45 degrees to the sides of the $1 m t r$ corridor that contains it.

Let length of base of rectangle =$x$ and have a height [ or width]of $h$.

Area of rectangle will equal $x h$.........$\left[1\right]$. From the sketch, it is seen that $\left[2 \sqrt{2} - x\right] = 2 a$ ,........ie, $a = \frac{\left[2 \sqrt{2} - x\right]}{2}$.........$\left[2\right]$.

Noting the small right angled triangle whose base is $a$ and whose height $h$.

Tan $\frac{\pi}{4}$ = $\frac{a}{h}$, and so from .......$\left[2\right]$, and after substituting for $a$ $a = \frac{\left[2 \sqrt{2} - x\right]}{2}$
$h$ = $\frac{\left[2 \sqrt{2} - x\right]}{2}$,[ since tan pi/4=1] so from ........$\left[1\right]$ area of rectangle= $\left[x\right] \frac{2 \sqrt{2} - x}{2}$

= $\sqrt{2} \left[x\right]$-${x}^{2} / 2$. Differentiating this expression, $\frac{d}{\mathrm{dx}} = \sqrt{2} - x$. This expression must equal zero for max/ min, i.e, $x = \sqrt{2}$. and substituting this value for $x$ into the expression for $h$ in terms of $x$ will give a value of $h$ = $\frac{\sqrt{2}}{2}$. The second derivative is negative = -$1$ [ which is negative whatever the value of $x$]. This confirms that the $x = \sqrt{2}$ will maximise the area of the rectangle.

I hope this was helpful and I believe it is correct.