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A:
Some comments about the question and your current solution.
Firstly, I’m surprised that you need to get a quadratic function from this problem to find the maximum. It should be possible to do this using only a linear function. Assuming you already know that the maximum of $f(x)=x^3$ is $x=2$, for instance, and that the $\max$ (maximum) of $g(x) =2x-x^2$ is somewhere between $1$ and $2$, we can “double” the values of the function at those points to get something nicer: $$h(x) = g(2x)-2g(x) = (2x-x^2)^2 -2(2x-x^2)=x^4-2x^3+2x^2-4x+1$$
Clearly it has its maximum at $x=2$, so you can get the maximum from the derivative $h'(2)$, but it is not so easy to find that. Taking the derivative, we have $h'(2)=-4$, so the maximum is $h(2)=17-8+1=18$. We can even get a closed form: $h'(x)=4(2x-x^2)-3(2x-x^2)=0$, so $h(x)=18=x^4$, meaning that $2$ is the maximum.
I don’t see why you want to get $A\leq\sin(B)$, though. You could do this without a quadratic equation: The line $y=x$ and the line $y=-x$ partition the space into a quadrant and you want the one that contains the max. Can you see why this works?
it doesn’t matter. And we’ll be the judge of that.
The point is, the rest of the world doesn’t have to be lost in the details. They don’t have to be watching every arrow, every loop, and every button press of the game to try to catch
