Can the angle take on any value?

Let's do the general case of a parallelogram with perimeter $P$ and area $A$ .

Label the side lengths $a = \frac{P}{4} + t$ and $b = \frac{P}{4} - t$ . The parallelogram has the correct perimeter for all values of $t$ , and the natural range for $t$ is $-\frac{P}{4} \leq t \leq \frac{P}{4}$ .

The area of the parallelogram depends on $t$ and the angle $\theta$ between two adjacent sides. $\begin{aligned} A &= ab \sin \theta \\ &= \left(\frac{P}{4} + t \right) \left( \frac{P}{4} - t \right) \sin \theta \\ &= \frac{1}{16} \left( P^2 - 16t^2 \right) \sin \theta \end{aligned}$ Solving for the angle, we get $\sin \theta = \dfrac{16A}{P^2 - 16t^2}$ . This has a minimum when the denominator is a maximum. That occurs when $t = 0$ , meaning we have a rhombus with $a = b = \frac{P}{4}$ . $\boxed{ \sin \theta \geq \frac{16A}{P^2} }$ .

Plugging in the values given, we get $\sin \theta \geq \dfrac{16(250)}{80^2} = \boxed{ \frac{5}{8} }$ . This corresponds to angles greater than $\sin^{-1} \left( \frac{5}{8} \right) = 38.7^\circ$ and less than $180^\circ - 38.7^\circ = 141.3^\circ$ .

Let $b$ be the total length of the base, $h$ the height, and $d = \frac{h}{\sin(\theta)}$ the length of the hypotenuse of the right triangle. From the problem statement, we can see that $bh = 250$ and $2(b+d) = 80$ . Using these equations, we can find an equation for $\sin(\theta)$ in terms of $h$ . $b+d = \frac{250}{h} + \frac{h}{\sin(\theta)} = 40$ $250 + \frac{h^2}{\sin(\theta)} = 40h$ $\frac{h^2}{\sin(\theta)} = 40h -250$ $\sin(\theta) = f(h) = \frac{h^2}{40h-250}$ Differentiate using the quotient rule and set to zero to find the minimum. $f'(h) = \frac{2h(40h-250) - 40h^2}{(40h-250)^2} = \frac{40h^2 -500h}{(40h-250)^2}$ Set the numerator to zero since the denominator can't be zero. $40h^2 - 500h = 20h(2h-25) = 0 \longrightarrow h = 0, \frac{25}{2}$ Clearly, $h\neq 0$ or else the area would be 0, so $h = \frac{25}{2}$ . Thus, the minimum of $\sin(\theta)$ is $f\left(\frac{25}{2}\right) = \frac{\left(\frac{25}{2}\right)^2}{40\left(\frac{25}{2}\right) - 250} = \frac{\frac{625}{4}}{250} = \frac{625}{1000} = \boxed{0.625}$

$let\space the\space sides\space be\space x,y$ $\Rightarrow x+y+x+y=80$ $\Rightarrow x+y=40..........[1]$ $\Rightarrow Area = xy\sin\theta=250$ $\Rightarrow \sin\theta=\dfrac{250}{xy}..........[2]$ $Squaring \space [1]$ $\Rightarrow x^2+y^2+2xy=1600$ $\Rightarrow xy=\dfrac{1600-(x^2+y^2)}{2}$ $Putting\space this\space in\space [2]$ $\sin\theta=\dfrac{500}{1600-(\red{x^2+y^2})}$ $\Rightarrow\sin\theta\space will\space be \space minimum\space when\space 1600-(\red{x^2+y^2})\space will \space be\space maximum$ $\Rightarrow\sin\theta\space will\space be \space minimum\space when\space \red{x^2+y^2}\space will \space be\space minimum$ $\red{x^2+y^2}=x^2+(40-x)^2\space\space\space\blue{\because x+y=40 \space from\space [1]}$ $=2x^2-80x+1600$ $Taking\space the\space first\space derivative\space test\space we\space get$ $x=20$ $\Rightarrow\sin\theta=\dfrac{5\cancel{00}}{16\cancel{00}-(4\cancel{00}+4\cancel{00})}$ $=\dfrac{5}{8}=\boxed{0.625}$

Let $x$ and $y$ be lengths of this parallelogram such that:

AREA: $2 \cdot \frac{1}{2}xy \cdot \sin \theta = 250 \Rightarrow \sin \theta = \frac{250}{xy}$ (i)

PERIMETER: $2x + 2y = 80$ (ii)

If $y = 40 -x$ from (ii), then substitution into (i) gives us $\sin \theta = f(x) = \frac{250}{x(40-x)} = \frac{250}{40x-x^2}$ (iii). Taking $f'(x) = 0$ yields:

$f'(x) = -\frac{250(40-2x)}{(40x-x^2)^2} = 0 \Rightarrow x = 20$

and the second derivative at $x = 20$ gives:

$f''(x) = 250[\frac{2(40-2x)^2}{(40x-x^2)^3} + \frac{2}{(40x-x^2)^2}] \Rightarrow f''(20) = \frac{1}{320} > 0$ (which is a global minimum).

Hence, $f_{MIN} = f(20) = \frac{250}{40(20)- 20^2} = \frac{250}{400} = \boxed{\frac{5}{8}}.$

平行四辺形の二組の辺をt,s　面積をSと置く。 2t+2s=80 tssinθ=250 の二つの式を連立するとsinθについてtまたはsで表現できる。 sinθ<1にそれを代入して平方完成すると、s=20でf(s)が最小の値をとる。 sで表されたsinθの式にs=20を代入して、答えは0.625