If a parabola is of the form $\frac{1}{8}$ ${(42x-15x^2)}$, what would be the optimised area a rectangle can have inside it if the bottom (width) of the rectangle is aligned on the X axis i.e the y isn't less than zero?
Edit: Thank you to everyone for your help!
Best Answer
The rectangle with maximum area will clearly have its upper vertices lying on the parabola (as otherwise we can expand the rectangle to a bigger one).
This means, due to symmetry, that the rectangle's midpoint will have the same $x$ coordinate as the parabola's vertex, that is, $\dfrac{7}{5}$.
The area of the rectangle will then be, depending on the half-length $a$ of the base of the rectangle, $f(a)=2a\cdot \left(\dfrac{42(7/5+a)-15(7/5+a)^2}{8}\right)$.
Differentiating we get $$f'(a)=-\dfrac{3}{20}(75a^2-49)$$ and thus $f'(a)=0\iff a=\sqrt{\dfrac{49}{75}}=\dfrac{7}{5\sqrt{3}}$, and this obviously must be a maximum (as one can check with the second derivative).
Notice that the width in this case would be $\dfrac{14}{5\sqrt{3}}\approx 1.616$, which is really similar to the value you mentioned gave a 'large' area.
Thus, the rectangle with maximum area will have the segment $\left[\dfrac{7}{5}-\dfrac{7}{5\sqrt{3}},\dfrac{7}{5}+\dfrac{7}{5\sqrt{3}}\right]$ as its base (with its upper vertices lying on the parabola).