Find the absolute min/max

f(t)=4cost+2sin2t, [0,pi/2]

What is ABSOLUTE MAX?

$\begin{array}{|rcll|} \hline f(t) &=& 4\cdot \cos(t) + 2\cdot \sin(2t) \qquad 0\le t\le \frac{\pi }{2}\\ f'(t) &=& 4\cdot[-\sin(t)] + 2\cdot \cos(2t) \cdot 2 \\ f'(t) &=& -4\cdot \sin(t) + 4\cdot \cos(2t) & | \quad \text{find min/max } f'(t) = 0 \\ 0 &=& -4\cdot \sin(t) + 4\cdot \cos(2t) & | \quad : 4 \\ 0 &=& - \sin(t) + \cos(2t) \\ \cos(2t) - \sin(t) &=& 0 & | \quad \cos(2t) = \cos^2(t)-\sin^2(t)\\ && & | \quad = 1-\sin^2(t)-\sin^2(t) \\ && & | \quad = 1-2\sin^2(t)\\ \cos(2t) - \sin(t) &=& 0 & | \quad \cos(2t) = 1-2\sin^2(t)\\ 1-2\sin^2(t) - \sin(t) &=& 0 & | \quad \cdot (-1) \\ -1+2\sin^2(t) + \sin(t) &=& 0 \\ 2\sin^2(t) + \sin(t) -1 &=& 0 & | \quad \text{substitute: } u = \sin(t)\\ 2u^2 + u -1 &=& 0 \\ u_{1,2} = \frac{ -1\pm \sqrt{(-1)^2-4\cdot2\cdot (-1)} } { 2\cdot 2 }\\ u_{1,2} = \frac{ -1\pm \sqrt{1+8} } {4 }\\ u_{1,2} = \frac{ -1\pm \sqrt{9} } {4 }\\ u_{1,2} = \frac{ -1\pm 3 } {4 }\\ u_1 &=& \frac{-1+3}{4} \\ u_1 & =& \frac12 \\ u_2 &=& \frac{-1-3}{4} \\ u_2 & =& -1 \\ t_1 &=& \arcsin(u_1) \\ t_1 &=& \arcsin(\frac12) \\ \mathbf{t_1 }& \mathbf{=} & \mathbf{30^{\circ} \text{ or } t_1 = \frac{\pi}{6} }\\ t_2 &=& \arcsin(u_2) \\ t_2 &=& \arcsin(-1) \\ t_2 &=&-90^{\circ} \text{ or } t_2 = -\frac{\pi}{2} & | \quad \text{no solution because } 0\le t\le \frac{\pi }{2}\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline f(t) &=& 4\cdot \cos(t) + 2\cdot \sin(2t) \\ f(30^{\circ}) &=& 4\cdot \cos(30^{\circ}) + 2\cdot \sin(2\cdot 30^{\circ}) \\ f(30^{\circ}) &=& 4\cdot \cos(30^{\circ}) + 2\cdot \sin(60^{\circ}) \\ f(30^{\circ}) &=& 4\cdot \frac{ \sqrt{3} } {2} + 2\cdot \frac{ \sqrt{3} } {2} \\ f(30^{\circ}) &=& 2\cdot \sqrt{3} + \sqrt{3} \\ f(30^{\circ}) &=& 3\cdot \sqrt{3} \\ f(30^{\circ}) &=& 5.19615242271 \\ \hline \end{array}$

Max or min at $\mathbf{t } \mathbf{=} \mathbf{30^{\circ} ( = \frac{\pi}{6} } ) \qquad f(30^{\circ}) = 5.196\dots \quad ?$
$\begin{array}{|rcll|} \hline y"(t) &=& -4\cdot \cos(t)-4\cdot\sin(2t)\cdot 2 \\ y"(t) &=& -4\cdot \cos(t)-8\cdot\sin(2t)\\ y"(30^{\circ}) &=& -4\cdot \cos(30^{\circ})-8\cdot\sin(2\cdot 30^{\circ})\\ y"(30^{\circ}) &=& -4\cdot \cos(30^{\circ})-8\cdot\sin(60^{\circ})\\ y"(30^{\circ}) &=& -4\cdot \frac{ \sqrt{3} } {2}-8\cdot\frac{ \sqrt{3} } {2}\\ y"(30^{\circ}) &=& -12\cdot \frac{ \sqrt{3} } {2} \\ y"(30^{\circ}) &=& -6\cdot \sqrt{3} & | \quad \text{maximum because } y''(30^{\circ}) < 0\\ \hline \end{array}$

Max at $\mathbf{t } \mathbf{=} \mathbf{30^{\circ} ( = \frac{\pi}{6} } ) \qquad f(30^{\circ}) = 5.196\dots \quad$