I have been working on a textbook's Optimization problem but the answer that I got does not match the textbook's answer. I would like to make sure I got it right (I can't find any mistakes on my solution), so I would like to ask for someone's help. I would appreciate that.
The problem:
A rectangle is located below a parabola, which is given by $$y = 3x- \frac{x^2}{2}$$ in such a way that its two superior vertexes are placed on the parabola and its two inferior vertexes are located on the $x$ axis. The left, inferior vertex, is placed on the point $(c,0)$. That said:
a) Show that the area of the rectangle can be represented by the equation $$A(c) = c^3 -9c^2 + 18c$$
b) Find the rectangle's height and width given that is has maximum possible area.c) What is that area?
Instead of typing the whole solution I will post an image with it (sorry for that but latex-ing it would take a lot of time!). You will find my solution below.
If that's helpful, the textbook's answer for b) and c) are $3-\sqrt{3} \times 3$ and $9 + 9\sqrt{3}$ respectively.
Thank you.
Best Answer
You made a slight error when trying to find the base:
$$c = 3-\sqrt{3} \implies b = 2(3-c) = \color{blue}{2\left[3-\left(3-\sqrt{3}\right)\right]} = 2\sqrt{3}$$
You forgot the $3$ in $(\color{blue}{3}-c)$ and found $b = 2c$ instead.
Addition: From here, using $h = 3$ as you found, you get
$$A = bh \iff A = 3\left(2\sqrt{3}\right) = 6\sqrt{3}$$
Try plugging in $c = 3-\sqrt{3}$ in $f(c)$:
$$f\left(3-\sqrt{3}\right) = \left(3-\sqrt{3}\right)^3-9\left(3-\sqrt{3}\right)^2+18\left(3-\sqrt{3}\right) = 6\sqrt{3}$$
Through confirmation, you can see this point coincides with the local maximum.