The hypotenuse of a right angle triangle measures 12 cm. What size angles would produce maximum... The hypotenuse of a right angle triangle measures 12 cm. What size angles would produce the maximum perimeter?
I got to point where I take the derivative and get $12(\cos\theta-\sin\theta)=0$, not sure where to go from there.
 A: Perimeter of a triangle
$$
P = a+b+12
$$
which needs to be maximized with constraint
$$
a^2+b^2=144
$$
Easiest way to do that is constructing Lagrangian using multipliers
$$
L = a+b+12-\lambda\left( a^2+b^2-144\right)
$$
so
$$
\frac {\partial L}{\partial a} = 1-2\lambda a=0 \\
\frac {\partial L}{\partial b} = 1-2\lambda b=0 \\
\frac {\partial L}{\partial \lambda} = a^2+b^2-144=0
$$
One can easily find that 
$$
a = b = \frac {12}{\sqrt 2}
$$
And therefore $\theta = \frac \pi 4$, since triangle is isosceles.
Note
One can also see that answer doesn't really depend on the value of hypotenuse, and problem can be formulated as "Find angles of the right angle triangle with given hypotenuse $c$ and maximum possible perimeter." All of those triangles will be isosceles.
A: Now you need to solve $\cos\theta-\sin\theta = 0$, i.e. $\cos\theta=\sin\theta$, so
$$\tan\theta=1$$
From here, since $0 < \theta < \frac{\pi}{2}$,
 $$\theta=\frac{\pi}{2}-\theta=\frac{\pi}{4}$$
To check that this is indeed the maximum, the second derivative test is needed.
$$\left. \frac{d}{d\theta}[12(\cos\theta-\sin\theta)]\right|_{\theta=\frac{\pi}{4}} = \left. -12(\sin\theta + \cos\theta)\right|_{\theta=\frac{\pi}{4}} = -12\sqrt{2}<0$$
Hence $\theta=\frac{\pi}{4}$ is the point of maximum.
A: Let's call the legs $x$ and $y$, and the perimeter $P$. So 
$$P=x+y+12$$
Draw  the triangle, and by the Pythagorean theorem you should be able to see that
$$
y=\sqrt{144-x^2}
$$
Now substitute back to get
$$P=x+\sqrt{144-x^2}+12
$$
Therefore, 
$$P'=1-\frac x{\sqrt{144-x^2}}$$
Now set this equal to $0$ and you should get $x$ to be $6\sqrt 2$. Now you have the hypotenuse and the adjacent leg, so
$$\cos \theta=\frac {6\sqrt 2}{12}$$
or
$$\cos \theta = \frac {\sqrt 2}2
$$
Therefore, $\theta=45^\circ$ or $\frac \pi 4$  
A: So, the perimeter will be $12(1+\cos\theta+\sin\theta)=12+12(\cos\theta+\sin\theta)$
Now, $\sin\theta+\cos\theta=\sqrt2\cos(\theta-\frac\pi4)$ which will be maximum if $\theta-\frac\pi4=2n\pi$ where $n$ is any integer 
So, $\theta=2n\pi+\frac\pi4$
As $0<\theta<\frac\pi2, \theta=\frac\pi4$
A: We want to maximize $12(\sin\theta+\cos\theta)$. So we want to maximize $\sin\theta+\cos\theta$. 
Equivalently, we want to maximize $(\sin \theta+\cos\theta)^2$. So we want to maximize $\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta$. 
So we want to maximize $\sin 2\theta$. This happens when $2\theta=\frac{\pi}{2}$.  
