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.

enter image description here

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.

  • 1
    $\begingroup$ $base=2\sqrt{3}$ $\endgroup$ – Aleksas Domarkas Jan 6 at 8:19

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.

  • $\begingroup$ thank you I've fixed that mistake. Now I have $2 \sqrt{3}$ as a base and $3$ as the rectangle's height, which gives me an area of $A = 6 \sqrt{3} \ m^2$. Am I good? Thanks for your help! $\endgroup$ – bru1987 Jan 6 at 9:16
  • $\begingroup$ Yes, that’s the correct answer! $\endgroup$ – KM101 Jan 6 at 9:24
  • $\begingroup$ thank you and have a great day =) $\endgroup$ – bru1987 Jan 6 at 9:26
  • 1
    $\begingroup$ Glad to have helped! :-) You may want to mark the answer as accepted if it has helped resolved the issue. $\endgroup$ – KM101 Jan 6 at 9:31

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