In the closed area limited by the graph $y = 4-x^2$, an isosceles triangle is inscribed. The triangle has its top angle in the origin and its base is parallel to the x-axis. Decide the triangle's maximal area.
My question is the following: Why do you draw the triangle as in the picture below and not in any other way?
Set the top vertex in the origin. "In the closed area limited by the graph" sounds a little bit sloppy formulated, because the graph itself does not limit any closed area. I guess they mean "limited by the graph and the $x$-axis". Then the base should be above the $x$-axis and below the line $y=4$. "Parallel to $x$-axis" - as it says. "Inscribed" means two other vertices on the parabola. Otherwise, no more restrictions. The base on the picture goes through $y=3$, but it is just a coincidence. You can draw it any other way, it is simply an example of a possible triangle (not likely the one that maximizes the area). When you vary the base all the possible ways between the $x$-axis and the line $y=4$ you'll get different triangles, and you are to find the one with the largest area.