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- #1

- Thread starter Jenny
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- Thread starter
- #1

- Feb 13, 2012

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Someone please help me with this question. I can't do and I have a calculus exam in the morning.

View attachment 762

Let suppose that is symply L=1 and set $\displaystyle \theta$ the angle between the two side of length 1. The area is...

$\displaystyle A = \sin \frac{\theta}{2}\ \cos \frac{\theta}{2}$ (1)

... and You can maximise A forcing to zero the derivative of (1)...

$\displaystyle \frac{d A}{d \theta} = \frac{1}{2}\ \{ \cos^{2} \frac{\theta}{2} - \sin^{2} \frac{\theta}{2}\} = 0$ (2)

The (2) is satisfied for $\displaystyle \theta = \frac{\pi}{2}$ and that means that...

$\displaystyle A_{\text{max}} = \sin \frac{\pi}{4}\ \cos \frac{\pi}{4} = \frac{1}{2}$ (3)

The triangle with langest area is the half of a square!...

Kind regards

$\chi$ $\sigma$

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- #3

Hello Jenny,

An alternate to

\(\displaystyle A(\theta)=\frac{L^2}{2}\sin(\theta)\) where \(\displaystyle 0<\theta<\pi\)

Now, equating the derivative to zero, we find:

\(\displaystyle A'(\theta)=\frac{L^2}{2}\cos(\theta)=0\,\therefore\,\theta=\frac{\pi}{2}\)

and so what is the length of $x$?

To ensure we have maximized the area, we see that:

\(\displaystyle A''(\theta)=-\frac{L^2}{2}\sin(\theta)\)

\(\displaystyle A''\left(\frac{\pi}{2} \right)=-\frac{L^2}{2}<0\)

Another approach would be to let $x$ be the base of the isosceles triangle, so we would need to express the altitude $h$ as a function of $x$ and $L$, and this can be accomplished by bisecting the triangle along the altitude to get two right triangles.

Can you find the altitude?