I have to find the max area of a rectangle in an inverted parabola with the equation y=3x(4-x). I expanded the equation to be 3x^2-12x+h to find where the corners lie on the x-axis. If i solve for h, will that give me my 2 corners that lie on the x-axis? Which is (2+/-(3-h)^0.5/2?? Im lost as to find what the x coordinate is.
2026-04-23 14:31:24.1776954684
Max possible area of a rectangle that has its corners on the inverted parabola
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Hint: I am assuming that you are referring to the rectangle inscribed between the x axis and the parabola. So, note that the rectangle will be symmetric about the axis of the parabola. Therefore, assuming the coordinates of one vertex, you can identify the same for the other one on the x axis. Having done this, use the equation describing the parabola to find the other two vertices (essentially find where the vertical lines from the aforementioned pair of vertices intersect). Thus you have 4 vertices, so you have the area. Now maximise over your chosen variable.