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I needed assistance checking a solution to a calculus problem.

Consider the graph of the function $f: [-\frac{\pi}{2}, \frac{\pi}{2}] \rightarrow \mathbb{R}$ given by $f(x) = \cos(x)$. Note that it is an arch shape. The problem asks for us to inscribe a rectangle in this arch, maximizing its area. In particular, we want its dimensions. The idea is that the "top" corners of the rectangle touch the graph and the "bottom" corners touch the $x$-axis. The two sides of the rectangle are parallel to the axes.

Attempted solution: We note that $f$ is an even function. To be more precise, $f(x) = f(y) \Rightarrow x=y$ or $x=-y$. The two $x$-coordinates of the upper corners of the rectangle, since they are on the same horizontal line, must therefore be additive inverses of each other. This implies, moreover, that the horizontal length of the rectangle is $2c$ where $c$ is the positive value of the pair. The height of the rectangle is of course $\cos(c)$

The area of the rectangle can be given by $A:(0, \frac{\pi}{2}) \rightarrow \mathbb{R}$ where $A(x) = (2x)(\cos x)$ and $x$ is the $x$-coordinate of the upper-right corner.

The function must be maximized. $A'(x) = 2(\cos(x)-x \sin(x))$. This vanishes and changes sign from positive to negative at $x \approx 0.8603335$ (note that numerical approximation suffices). This is thus the maximum of $A(x)$ within its domain.

$2(0.8603335) = 1.720667$ and $\cos(0.8603335) \approx 0.652$. The dimensions of the rectangle are therefore $\approx 1.720667 \times 0.652$.

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  • $\begingroup$ "This is thus the maximum of $A(x)$." -- not quite. it's the x coordinate where the maximum occurs. $\endgroup$ – symplectomorphic Jan 13 '16 at 9:51
  • $\begingroup$ Your work is very correct, indeed ! But, what is the question ? $\endgroup$ – Claude Leibovici Jan 13 '16 at 9:53
  • $\begingroup$ @ClaudeLeibovici I just needed to know if it's correct. I believe these sorts of questions are allowed, if I'm not mistaken, where the OP asks simply to verify proofs/solutions. $\endgroup$ – MathematicsStudent1122 Jan 13 '16 at 9:54
  • $\begingroup$ You are right and your approach is very rigorous. The only thing you could develop is the manner you solved $A'(x)=0$. This could be a nice complement to the solution. $\endgroup$ – Claude Leibovici Jan 13 '16 at 9:57
  • $\begingroup$ @ClaudeLeibovici In fact, I had simply used a calculator to determine that. Is there a more rigorous manner in which the zero could have been determined? $\endgroup$ – MathematicsStudent1122 Jan 13 '16 at 10:01
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There is a little caveat here: Assume you had been asked to determine the rectangle with smallest area. Proceeding as you did you would have arrived at the same rectangle $1.720667\times0.652$. What went wrong here?

You want to find the maximum of the function $x\mapsto A(x)$ on the interval $I:=\bigl[0,{\pi\over2}\bigr]$. This maximum is taken at one of the endpoints of this interval, or at an interior point where $A'(x)=0$. Since $A(0)=A\bigl({\pi\over2}\bigr)=0$ (you should have checked this!), and $A(x)$ is positive in the interior of $I$ the point we are looking for is indeed the point you have found.

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