How to find the least in $E^{\circ}=\frac{5S^g}{162}+\frac{C^\circ}{50}+\frac{2\pi^2}{360}\textrm{rad}$? The problem is as follows:
Let $S^{\circ}$, $C^{g}$ and $R\,\textrm{rad}$ the measures of a positive angle in sexagesimal, centesimal and radian degrees respectively such as:
$$E^{\circ}=\frac{5S^g}{162}+\frac{C^\circ}{50}+\frac{2\pi^2}{360R}\textrm{rad}$$
Using this information find the least $E$.
The alternatives in my book are as follows:
$\begin{array}{ll}
1.&\textrm{6}\\
2.&\textrm{8}\\
3.&\textrm{10}\\
4.&\textrm{12}\\
\end{array}$
How exactly should I find the least angle here?.
So far what I remember is that the proportions between each unit are the same in the sense of:
$\frac{S}{360}=\frac{C}{400}=\frac{R}{2\pi}$
Using this information you may find a relationship between the variables.
Plugin these in the above equation I'm getting:
However I find confusing why the degree symbols for $S$ and $C$ appears swapped then I figured that what it might be intended could be that the digits for the angle are represented by the letter but the degree is contrary to what its unit mentions.
For $\frac{5S^g}{162}$:
$\frac{5S^g}{162}\times\frac{360^\circ}{400^g}=\frac{1S}{36}^{\circ}$
For $\frac{C^\circ}{50}$:
Since;
$\frac{S}{360}=\frac{C}{400}$
$C=\frac{10}{9}S$
$\frac{C^\circ}{50}=\frac{10S^\circ}{9}\times \frac{1}{50}=\frac{1S^\circ}{45}$
For: $\frac{2\pi^2}{360R}\textrm{rad}$
Since this case $R$ matches with the degree symbol $\textrm{rad}$ then I would apply the equation:
$\frac{S^{\circ}}{180}=\frac{R\rad}{\pi}$
$R \textrm{rad}=\frac{S^\circ \pi}{180}$
$\frac{2\pi^2}{360R}\textrm{rad}=\frac{2\pi^2 \times 180}{360S^\circ \pi}=\frac{\pi}{S^\circ}$
Then:
$E^{\circ}=\frac{5S^g}{162}+\frac{C^\circ}{50}+\frac{2\pi^2}{360R}\textrm{rad}$
$E^{\circ}=\frac{1S}{36}^{\circ}+\frac{1S^\circ}{45}+\frac{\pi}{S^\circ}$
$E^{\circ}=\frac{S^2+20\pi}{20S}$
Then I assume that this latter expression must me minimized but would it be used calculus for this?
If so:
$E^{\circ}\,'=\frac{1}{20}-\frac{\pi}{S^2}$
Equating this to zero:
$E^{\circ}=2\sqrt{5\pi}$
But I don't know if this is what it was intended?.
Can someone help me here?. I'm stuck, why it doesn't matches any of the alternatives given or could it be that I made a mistake?. Please help.
 A: While it is true that $R$ rad $=\dfrac {S^\circ \pi}{180^\circ} $, it is not true that $\dfrac {2\pi^2}{360R}$rad $= \dfrac {2\pi^2 \times 180^\circ}{360S^\circ\pi}$; the latter is actually equal to $\dfrac {2\pi^2}{360R \text{ rad}}$. To resolve this, we consider the numerical value of $R$ and the conversion of rad to degrees (sexagesimal) seperately:
$$R = \frac {S \pi}{180}, \quad 1 \text{ rad }=\frac {180}{\pi}^\circ$$
Now we have:
$$\frac {2\pi^2}{360R} \text{ rad } = \frac {2\pi^2 \times 180}{360S\pi} \times \frac {180^\circ}{\pi}=\frac {180^\circ}{S}$$
$$E^\circ = \frac{5S^g}{162}+\frac{C^\circ}{50}+\frac{2\pi^2}{360R}\textrm{rad} = \frac {S^\circ}{36} + \frac {S^\circ}{45} + \frac {180^\circ }S = \left(\frac S{20}+\frac{180}S\right)^\circ$$
To minimize this we may use calculus, or better yet, the 2-term form of AM $\ge$ GM:
$$a+b\ge 2\sqrt{ab}$$
$$\frac S{20}+\frac{180}S\ge2\sqrt{\frac S{20}\times \frac {180}S} =6$$
Hence the minimum of $E^\circ$ is $6^\circ$ (which occurs at $S = 60$).
