Which calculus method should I use for this investigation? What does it mean choose a suitable length for the cross section?

Question

Your task is to determine which shape maximises the volume of water that can be held by a gutter of a fixed length of material. choose a suitable fixed length for the cross section and explore different shapes. You should consider a rectangular, triangular and round gutter as well as at least 2 other shapes. Determine the dimensions and cross-sectional area that maximises the water held by the gutter for each shape. Generalise each shape for a length L and make use of calculus to determine the dimensions of the shape and associated cross-sectional area that optimises the water held by the gutter.

what mathematical method do I approach this question with what constraints do I use? My peers all are using different methods and I don't which one to use some are using a perimeter and some are using a surface area which one is easier I have been able to do some sort of working out for the rectangular gutter and triangle however they are not coherent enough to hand in to the teacher. Furthermore, I cant get started with the semicircle gutter. Does anyone know how to go about this investigation?

btw my extent of knowledge is chain rule, quotient rule, power rule and second derivative so any math beyond that is not required for this.

Answer 1

I think your problem is to minimize the cross sectional perimeter for a fixed cross sectional area in order to minimize the material used in construction. Take for example a square with sides equal to 1. The area is 1 and the perimeter is 3. Now consider a half circle with area $A=\frac{\pi}{2} r^2 = 1$ and $r=\sqrt{\frac{2}{\pi}}$. The length of the perimeter is $\pi r=\sqrt {2\pi}=2.5066$. The circle is more optimal than a square. Now take an equilateral triangle with base b and height $h=\frac{\sqrt 3 b}{2}$. $A=\frac{\sqrt 3 b^2}{4}= 1$ and $b=\frac{2}{\sqrt {\sqrt 3}}$ and the perimeter is $\frac{4}{\sqrt {\sqrt 3}}=3.0394$. So in ranking optimal shapes we find circle ranks first, square ranks second and equilateral triangle ranks third.

Answer 2

If I understand correctly, you want to hold the perimeter of the cross section constant while maximizing the cross sectional area. It looks like you were asked to consider regular n-gons.

A regular n-gon can be broken up into $n$ Isosceles triangles with base = $L/n$.

The apothem will be $a\tan (\pi/n)=(1/2)(L/n)\implies a=\frac{L}{2n}\cot(\pi/n)$

Area is half the apothem times the perimeter: $A=(1/2)ap$

So $A(n)=\frac{L^2}{4n}\cot(\pi /n)=\frac{\pi L^2\cos(\pi /n)}{4\pi n(\sin \pi /n)}$

Tangent is monotonically increasing on the interval $ (0,\pi/2)$, so its reciprocal $\cot(x)$ is monotonically decreasing. Since $n$ is increasing, the argument for the cotangent is decreasing. Together these imply that the area increases monotonically with increasing $n$.

A circle is a regular polygon with infinite sides, so

$\lim_{n\to \infty} A(n)=\frac{L^2}{4\pi}$

You can prove that from $A(n)$ using the fact that $\lim_{x\to 0} (\sin x)/x =1$ and $\lim _{x \to 0} \cos x=1$

A more thorough consideration considers arbitrary plane curves, not just regular polygons, but that requires the calculus of variations. This approach considers only some potential cases but only requires you to use limits.