I came across this question regarding the maximum of the Lemur population.
We are supposed to do this without the use of any graphing software. I have differentiated the function but it still remains too complex to be solved.
This is what I got.
$$\frac{dy}{dx}=(x^2-7x+4)\cos(\frac{\pi x}{4})\frac{\pi}{4}+\sin(\frac{\pi x}{4})(2x-7)$$
My thought process was that within the interval $[0,2]$, $2x-7<0, \sin(\frac{\pi x}{4})>0, \cos(\frac{\pi x}{4})>0.$ The roots of $x^2-7x+4$ within the interval is $\frac{7-\sqrt{33}}{2}$. $\frac{dy}{dx}>0$ initally, but then at some value before $\frac{7-\sqrt{33}}{2}$, $\frac{dy}{dx}<0$, which means that the function is strictly decreasing afterwards. To find the minimum, one has to just plug in the extreme values to find $f(0)=20, f(2)=14$, so the minimum occurs when $x=2$.
I realise the same reasoning does not apply to find the maximum. Desmos says that the maximum occurs $0.303 \neq \frac{7-\sqrt{33}}{2}$. If I equate $\frac{dy}{dx}=0$, then I get that $x$ must satisfy this equation:
$$\tan(\frac{\pi x}{4})=\frac{\pi}{4}[\frac{x^2-7x+4}{7-2x}]$$
Which doesn't seem solvable by hand.
I've taught about the hint the question gave, but I don't have any ideas. Could someone give me a hint?