# Find the maximum dimensions that will strengthen rod

I did a quiz and we got the following question:

A cylindrical rod has a diameter of 2cm and on the cylinder is an isosceles triangle. The strength of the rod is proportional to the product of the base and the height squared. What are the maximum dimensions that will strengthen the cylinder.

We also got the following image:

I would really like to know the solution to this problem because no one in my class got the problem or had a thorough understanding of how to go about solving the problem.

Thanks a lot!

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I think you probably intended to ask for the dimensions that will maximally strengthen the rod? – joriki Mar 28 '13 at 20:19
@joriki Yea, I intended that – Jeel Shah Mar 28 '13 at 20:21
So ... fit the biggest isoceles triangle in a circle of fixed diameter? – Jacob Mar 28 '13 at 20:22
Or is the strength proportional to $bh^2$ and not $(bh)^2$? – Jacob Mar 28 '13 at 20:23
There's an edit button underneath the question. – joriki Mar 28 '13 at 21:00

Consider a circle of radius $R$ with an isosceles triangle inscribed, with base $b$ and height $h$. Draw lines from the center of the circle to the endpoints of the base to make another isosceles triangle. Draw the altitude of this triangle to the base to produce a right triangle having legs of $h-r$ and $b/2$, and hypotenuse $R$. Then

$$(h-R)^2 + \frac{b^2}{4} = R^2$$

from which you may deduce that

$$b=2 \sqrt{2 R h -h^2}$$

$$f(h) = b h^2 = 2 h^2 \sqrt{2 R h -h^2}$$

Of course, take $f'(h)=0$ and solve for $h$. You will get a linear equation with one solution. Proving that this solution is a max means evaluating $f''(h)$, which is a bit messy.

I will leave the mess to you. I get for the max strength:

$$f_{max} = \frac{50 \sqrt{5}}{27} R^3$$

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The answer should be $f_{max} = \frac{27}{4} R^4$. Take your $b$ into $f$ for $f=4 h^3 (2R -h)$. I think you missed a $\mbox{ }^2$ somewhere because the units of $f$ should be of $R^4$. – ja72 Mar 28 '13 at 21:00
@ja72: see the comment in the problem. The strength is proportional to $b h^2$, not $(b h)^2$ as you have it. The $R^3$ is correct. – Ron Gordon Mar 28 '13 at 21:01
oh! I see now. Thanks. I deleted my answer and I am officially endorsing this one. – ja72 Mar 28 '13 at 21:04

The radius of the cross section of the cylinder is equal to 1 cm.

If we draw three lines connecting the vertices of the triangle with the centre, we get one additional isosceles triangle with two sides 1 cm, and one side being the base of the bigger isosceles triangle.

Let the base be side AB, the top vertex be C and the centre of the circle be O.

$\angle{AOB} = 2\angle{ACB}$

Let $\angle{AOB} = \theta$

Then, $AB = 2\times r\times\sin\dfrac{\theta}{2} = 2\sin\dfrac{\theta}{2}$

The height (say h) of the triangle becomes:

h = 'Base to centre' + r

$H = r\cos\dfrac{\theta}{2} + r$

You want to maximise $bh^2$, so, you need to maximise:

$$\left(2\sin\dfrac{\theta}{2}\right)\left(\cos\dfrac{\theta}{2} + 1\right)^2$$

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If we denote height by $x$, then from Pythagorean theorem we obtain a formula for base $$y(x)=2\sqrt{r^2-(x-r)^2}=2\sqrt{2xr-x^2}$$ where $r$ is the radius of cylinder. Thus the required strength value will be $$f(x)=x^2*y(x)=2x^2\sqrt{2xr-x^2}$$ To maximise this value you'll need to differentiate it and look at function values in critical points (where derivative becomes 0 or is not defied). Also, take into account that $0<x<2r$.

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