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

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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

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Answer 1 · 2018-06-28T21:14:19

The rectangle will have a length of $\sqrt{2}$ $sqrt2$ , a width of $\frac{\sqrt{2}}{2}$ $sqrt2/2$ and thus area of $1 m^{2}$ $1m^2$ enter image source here

Explanation:

From the sketch it is seen that the length AB = BC = $\sqrt{2}$ $sqrt2$ [ 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 $1 m t r$ $1 mtr$ corridor that contains it.

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

Area of rectangle will equal $x h$ $xh$ ......... $[1]$ $[1]$ . From the sketch, it is seen that $[2 \sqrt{2} - x] = 2 a$ $[2sqrt2-x]=2a$ ,........ie, $a = \frac{[2 \sqrt{2} - x]}{2}$ $a = [[2sqrt2-x]]/2$ ......... $[2]$ $[2]$ .

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

Tan $\frac{π}{4}$ $pi/4$ = $\frac{a}{h}$ $a/h$ , and so from ....... $[2]$ $[2]$ , and after substituting for $a$ $a$ $a = \frac{[2 \sqrt{2} - x]}{2}$ $a=[[2sqrt2-x]]/2$
$h$ $h$ = $\frac{[2 \sqrt{2} - x]}{2}$ $[[2sqrt2-x]]/2$ ,[ since tan $\frac{π}{4} = 1]$ $pi/4=1]$ so from ........ $[1]$ $[1]$ area of rectangle= $[x] \frac{2 \sqrt{2} - x}{2}$ $[x][2sqrt2-x]/2$

= $\sqrt{2} [x]$ $sqrt2[x]$ - $\frac{x^{2}}{2}$ $x^2/2$ . Differentiating this expression, $\frac{d}{d x} = \sqrt{2} - x$ $d/dx= sqrt2-x$ . This expression must equal zero for max/ min, i.e, $x = \sqrt{2}$ $x=sqrt2$ . and substituting this value for $x$ $x$ into the expression for $h$ $h$ in terms of $x$ $x$ will give a value of $h$ $h$ = $\frac{\sqrt{2}}{2}$ $sqrt2/2$ . The second derivative is negative = - $1$ $1$ [ which is negative whatever the value of $x$ $x$ ]. This confirms that the $x = \sqrt{2}$ $x= sqrt2$ will maximise the area of the rectangle.

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