Find the absolute minimum and absolute maximum values of \( f \) on the given interval, \[ f(t)=16 \cos t+8 \sin 2 t, \quad[0, \pi / 2] \] absolute minimum absolute maximum

To find the absolute minimum and maximum of the function \( f(t) = 16 \cos t + 8 \sin(2t) \) on the interval \( [0, \frac{\pi}{2}] \), we start by finding the critical points by taking the derivative of \( f(t) \) and setting it to zero. 1. **Calculate the derivative:** \[ f'(t) = -16 \sin t + 16 \cos(2t) \] Using the double angle identity \( \cos(2t) = 2\cos^2(t) - 1 \): \[ f'(t) = -16 \sin t + 16(2\cos^2(t) - 1) = 32 \cos^2(t) - 16 \sin t - 16 \] 2. **Set the derivative to zero to find critical points:** \[ 32 \cos^2(t) - 16 \sin t - 16 = 0 \] We can express \( \cos^2(t) \) as \( 1 - \sin^2(t) \): \[ 32(1 - \sin^2(t)) - 16 \sin t - 16 = 0 \] Simplifying: \[ 32 - 32\sin^2(t) - 16\sin t - 16 = 0 \quad \Rightarrow \quad -32\sin^2(t) - 16\sin t + 16 = 0 \] Multiply by -1: \[ 32\sin^2(t) + 16\sin t - 16 = 0 \] 3. **Use the quadratic formula to solve for \(\sin(t)\):** \[ \sin(t) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 32 \cdot (-16)}}{2 \cdot 32} \] \[ = \frac{-16 \pm \sqrt{256 + 2048}}{64} = \frac{-16 \pm \sqrt{2304}}{64} \] \[ = \frac{-16 \pm 48}{64} \] Hence, we find two solutions: \[ \sin(t) = \frac{32}{64} = \frac{1}{2} \quad \text{and} \quad \sin(t) = \frac{-64}{64} = -1 \] Only \(\sin(t) = \frac{1}{2}\) is in the interval \([0, \frac{\pi}{2}]\), which gives us \( t = \frac{\pi}{6} \). 4. **Evaluate \( f(t) \) at the endpoints and the critical point:** - At \( t = 0 \): \[ f(0) = 16 \cos(0) + 8 \sin(0) = 16 \] - At \( t = \frac{\pi}{6} \): \[ f\left(\frac{\pi}{6}\right) = 16 \cos\left(\frac{\pi}{6}\right) + 8 \sin\left(\frac{\pi}{3}\right) = 16 \cdot \frac{\sqrt{3}}{2} + 8 \cdot \frac{\sqrt{3}}{2} = 12\sqrt{3} \] - At \( t = \frac{\pi}{2} \): \[ f\left(\frac{\pi}{2}\right) = 16 \cos\left(\frac{\pi}{2}\right) + 8 \sin(\pi) = 0 \] 5. **Compare the values:** - \( f(0) = 16 \) - \( f\left(\frac{\pi}{6}\right) = 12\sqrt{3} \approx 20.78 \) - \( f\left(\frac{\pi}{2}\right) = 0 \) Thus, the absolute minimum value is \( f\left(\frac{\pi}{2}\right) = 0 \) and the absolute maximum value is \( f\left(\frac{\pi}{6}\right) = 12\sqrt{3} \). **Absolute minimum: 0** **Absolute maximum: \( 12