## Optimising Trigonometric Function Example

What is the maximum length of a beam that can be squeezed around a corner?
Suppose we have a corridor with a corner. End side of the corner has a width of 5m and the other has a width of 3m.
From the diagram

$L=x+y=\frac{5}{cos \theta}+\frac{3}{sin \theta}$

Differentiating with respect to
$\theta$
and setting equal to zero gives
$\frac{dL}{d \theta} = \frac{d}{d \theta} (\frac{5}{cos \theta}+\frac{3}{sin \theta})=5 sec \theta tan \theta -3 cosec \theta cot \theta =0$

Hence
$5 sec \theta tan \theta =3 cosec \theta cot \theta$

$5 \frac{sin \theta}{cos^2 \theta} =3 \frac{cos \theta}{ sin^2 \theta}$

$\frac{sin^3 \theta}{cos^3 \theta} =\frac{3}{5} \rightarrow tan \theta = \sqrt{3}{\frac{3}{5}}$

But
$sec \theta = \sqrt{1+tan^2 \theta}=\sqrt{1+(\frac{3}{5})^{3/2}}$

and
$cosec \theta = \sqrt{1+cot^2 \theta}=\sqrt{1+(\frac{5}{3})^{3/2}}$

Hence
$L=5 \sqrt{1+(\frac{3}{5})^{3/2}} + 3 \sqrt{1+(\frac{5}{3})^{3/2}}$
.
From the graph the function is obviously a minimum.

The function here represents the distance between walls and the minimum distance is the maximum length of a beam that can pass through.

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