What is maximum value of $$f(t)=16\cos t \cdot \cos 2t \cdot\cos 3t \cdot \cos6t$$
My Approach:
$1. \;\;$Directly $t=0$ gives me maximum value of $16$.
$2.\;\;$ Converted it into $f(t)=\dfrac{\sin(4t)\sin(12t)}{\sin (t)\sin(3t)}\;\;$ but couldn't proceed further from this step.
My Doubts: $1.\;$Can we get maximum value without putting $t=0\;?$
$2.\;$ How can i proceed further using second method to get maximum value?
$3.\;$ Is there other way to solve this problem ?
On
I think you did not answer the interesting part of the problem: why is $t=0$ a maximum of $f$? Wang YeFei did answer this.
To answer your doubts/questions:
On
Answering your question...
Yes, but it would involve setting $f'(t)=0$ - you can use logarithmic differentiation$^1$ to get the derivative or use the product rule multiple times.
Same as 1, but now you'd have to involve the quotient rule AND the product rule to get things moving.
You could graph the equation and see where the maxima are by inspection. Desmos or Geogebra would be a huge help here.
$^1$ To find the logarithmic derivative, we would have...
$$f(t) = 16 \cos t \cos 2t \cos 3t \cos 6t \\ \ln f(t) = \ln 16 + \ln \cos t + \ln \cos 2t + \ln \cos 3t + \ln \cos 6t \\ \frac {f'(t)}{f(t)} = 0 + \frac {1}{\cos t}(-\sin t) + \frac {1}{\cos 2t}(-2\sin 2t) + \frac {1}{\cos 3t}(-3\sin 3t) + \frac {1}{\cos 6t}(-6\sin t) \\ f'(t) = f(t) (-\tan t -2 \tan 2t - 3 \tan 3t - 6 \tan 6t) \\ f'(t) = 16 \cos t \cos 2t \cos 3t \cos 6t(-(\tan t +2 \tan 2t + 3 \tan 3t + 6 \tan 6t))$$ Setting $f'(t) = 0$, you would have either $$16 \cos t \cos 2t \cos 3t \cos 6t = 0$$ or $$-(\tan t +2 \tan 2t + 3 \tan 3t + 6 \tan 6t)= 0$$
Observe that $\forall t, f(t) \le 16$ and $f(t) = 16$ when $t = 0$. This shows $16$ is the max value of $f$. All you need is $\cos(kt) \le 1$ for any $k,t$.