What is the Range of $5|\sin x|+12|\cos x|$ What is the Range of $5|\sin x|+12|\cos x|$ ?
I entered the value in desmos.com and getting the range as $[5,13]$.
Using $\sqrt{5^2+12^2} =13$, i am able to get maximum value but not able to find the minimum.
 A: If $f(x) = 5|\sin x| + 12 |\cos x|$, then
\begin{align*}
f(x) &= \sqrt{f(x)^2} \\
&= \sqrt{25 \sin^2 x + 144 \cos^2 x + 60 |\sin x \cos x|} \\
&= \sqrt{25 + (144 - 25) \cos^2 x + 60 |\sin x \cos x|} \\
&\ge 5
\end{align*}
with equality obtained when $\cos x = 0$.
A: The four quadrants give the four combinations of signs of $\sin$ and $\cos$.  Let us work initially in the first quadrant, where both functions are positive.  We can then remove the absolute value signs, take a derivative, and set to zero.
$$\frac d{dx}(5 \sin x + 12 \cos x)=5\cos x - 12 \sin x$$
This is zero when $\tan x=\frac 5{12}$, giving the maximum you found.  The minimum must then come at one end of the interval, and if you check $x=0, \frac \pi 2$ you find the minimum at $\frac \pi 2$, which is $5$.  You can do the same in the other four quadrants, flipping the signs of $\sin x$ and $\cos x$ as required, and find that the minimum is $5$ again.
A: Without loss of generality we may assume $0\le \theta\le\pi/2$ so $\cos \theta$ and $\sin \theta$ are positive.
You can solve this with calculus (set the derivative equal to zero and solve for $\theta$), but here is a geometric way to think about the problem which avoids calculus.
The level curves of the scalar function $f(x,y)=5y+12x$ are the parallel lines
$$ 5y+12x=C $$
for various real values $C$. One may "parametrize" such lines by using a perpendicular line through the origin, namely the line $y=\frac{5}{12}x$. Since the slope is less than $1$, the perpendicular line is tilted towards the $x$-axis, so the "first" of level curves to intersect the unit circle arc $0\le\theta\le\frac{\pi}{2}$ does so at the point $(0,1)$, which corresponds to an output of $f(0,1)=5$. The "last" of the level curves to intersect the arc occurs where the perpendicular line intersects it, so solve $x^2+(\frac{5}{12}x)^2=1$ to get $x=\frac{12}{13}$ and $y=\frac{5}{13}$, which yields a maximum of $f(\frac{12}{13},\frac{5}{13})=13$.
Therefore the range is $[5,13]$.
A: Another possible approach.
For the first quadrant: $5\sin(x) + 12\cos(x) = 13\sin(x + \arccos(\frac{5}{13}))$. Can follow from there for the rest of the quadrants. Can also try alternative forms for arguments in order to adapt to the values of $x$.
