A two-dimensional solid is made by alternatin | vedclass

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(A) The packing fraction $(PF)$ is defined as the ratio of the area of the circles to the area of the rectangle.$PF = \frac{\pi a^{2} + \pi b^{2}}{(2a

Vedclass · Accepted Answer

$0.41$

Vedclass · Answer

(A) The packing fraction $(PF)$ is defined as the ratio of the area of the circles to the area of the rectangle.$PF = \frac{\pi a^{2} + \pi b^{2}}{(2a + 2b)(2a)} = \frac{\pi(a^{2} + b^{2})}{4a(a + b)}$Dividing by $a^{2}$,we get $PF = \frac{\pi(1 + r^{2})}{4(1 + r)}$,where $r = \frac{b}{a}$.To minimize the packing fraction,we differentiate $PF$ with respect to $r$ and set it to zero:$\frac{d(PF)}{dr} = \frac{\pi}{4} \left[ \frac{(1+r)(2r) - (1+r^{2})(1)}{(1+r)^{2}} \right] = 0$$2r + 2r^{2} - 1 - r^{2} = 0$$r^{2} + 2r - 1 = 0$Using the quadratic formula $r = \frac{-B \pm \sqrt{B^{2} - 4AC}}{2A}$:$r = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$Since $r$ must be positive,$r = \sqrt{2} - 1 \approx 1.414 - 1 = 0.414$.Thus,the value is closest to $0.41$.

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