Question

Question deleted.

Answer 1

my guess would be a circle

Answer 2

what do you mean?

Answer 3

oh i see you asked a different question, which cannot really be answered without knowing the missing part up top,
on the other hand, the area is \(8600=\pi r^2\) making \(r=52.32\) approximately so
\[P=2\pi r = 2\pi \times 52.32\] whatever that is

Answer 4

that is ignoring the missing part up top
if you want exact without a deciimal
\[8600=\pi r^2\iff r=10\sqrt{\frac{86}{\pi}}\] and \[C=20\sqrt{86\pi}\]

Answer 5

the question i have is how do I minimize the perimeter of the circle given on the website. when i have the cross-sectional as 8600.

Answer 6

Minimizing the perimeter will give the minimum surface area, in turn giving the minimum amount of material to use.

@Phayeth I believe we've already resolved the question of the perimeter needed for the rectangular and triangular gutters, but I suppose it wouldn't hurt to repost that info here.

Rectangular X-section:

Solving for \(x\), you have \(x=\dfrac{8600}{y}\), and so the perimeter is
\[P=\frac{8600}{y}+2y\]
Differentiating, you get
\[\frac{dP}{dy}=-\frac{8600}{y^2}+2\]
and solving for the critical points:
\[\begin{align*}0&=-\frac{8600}{y^2}+2\\
\frac{8600}{y^2}&=2\\
4300&=y^2\\
y&=10\sqrt{43}
\end{align*}\]
which in turn gives \(x=20\sqrt{43}\). The perimeter is then \(40\sqrt{43}\).

Triangular X-section:

We determined the optimal angle first:
\[\frac{1}{2}x^2\sin\theta=8600~~\iff~~x=\sqrt{\frac{17200}{\sin\theta}}\]
Substituting into the perimeter equation,
\[P=2\sqrt{\frac{17200}{\sin\theta}}\]
and differentiating,
\[\frac{dP}{d\theta}=-\sqrt{17200}(\sin\theta)^{-3/2}\cos\theta\]
Setting equal to 0, and knowing that \(0\le\theta\le\pi\), you find \(\theta=\dfrac{\pi}{2}\text{ rad}=90^\circ\) to be the optimal angle, which means you have a 45-45-90 triangle. This in turn tells you that \(\dfrac{h}{x}=\dfrac{1}{\sqrt2}\) and \(\dfrac{h}{b/2}=1\).
\[\frac{h}{x}=\dfrac{1}{\sqrt2}~~\iff~~h=\frac{x}{\sqrt2}\\
\frac{h}{b/2}=1~~\iff~~b=\frac{2x}{\sqrt2}\]
So the area solely in terms of \(x\) is
\[\frac{1}{2}\left(\frac{2x}{\sqrt2}\right)\left(\frac{x}{\sqrt2}\right)=\frac{1}{2}x^2=8600\]
which gives \(x=\sqrt{17200}=20\sqrt{43}\), and hence a perimeter of \(40\sqrt{43}\).