I left a couple of problems (and their source) in an earlier comment.
Here are a couple more:
I'm not sure about the original sources for either of the following, but they should fit some interpretation of non-routine problems involving quadratic functions.
- I gave an example in an answer to The philosophy of change of variables (MSE 226343) that we might as well update for next year:
$$(x-1)(x-2)(x-3)(x-4) = 2016$$
Observe $(x-1)(x-4) = x^2 - 5x + 4$ and $(x-2)(x-3) = x^2 - 5x + 6$. If you use a $u$ substitution for the former product, then the problem becomes $u(u+2) = 2016$, which is quadratic; so you can solve for $u$ and then solve for $x$ (though you might pick a nicer number than $2016$ to set this problem up!).
An alternative method of solution is to observe symmetry around $x = 3/2$; use a substitution to shift the graph $3/2$ to the left and you end up with a function symmetric about $x= 0$, i.e., even. A similar substitution reduces the problem back to a routine quadratic situation, and, in fact, the substitutions in both of the above approaches end up being equivalent.
- Suppose you have $60m$ of fence and want to build a rectangular fence that maximizes the enclosed area. What are its dimensions? Okay, you have probably seen this problem enough times for it not to be considered non-routine. But how about a slight variation: Suppose you have access to a very long wall that can serve as one side of the rectangular fence. Again, build a rectangular enclosure that maximizes the enclosed area. Guess the dimensions, and then confirm whether your intuition was correct.
The first problem is, indeed, solved by a square: Each side is $15m$. You can tackle this by labeling adjacent sides as $x$ and $30-x$, so the area is $x(30-x)$; maximizing this downward facing parabola amounts to findings its vertex, which is why this is a quadratic problem.
An alternate solution, equivalent to the AM-GM inequality: The area of the $15m \times 15m$ rectangular enclosure is $225 m^2$; if we alter the dimensions, it makes use of additive inverses. That is to say, if we change one of the sides to $15+x$, then the adjacent side changes to $15-x$ in order to maintain a perimeter of $60$ (all measured in meters). But $(15+x)(15-x) = 225- x^2$ is clearly maximized when $x=0$ (i.e., in the aforementioned square case). In particular, a change to one of the sides by $x$ will decrease the enclosed area by $x^2$.
For the modified version, you now have access to the wall. So you can use all $60m$ to build the remaining three sides of your rectangular closure. Clearly, then, we end up with a square of dimensions $20m \times 20m$ with total area $400 m^2$. Right?
(Wrong. Why? Coming up with an "intuitive" explanation for what is the right answer perhaps qualifies as falling outside of the routine canon around quadratic function problems.)