Find the rectangle of maximum area if its vertices are at $(0,0)$, $(x,0)$, $(x,\cos x)$, $(0,\cos x)$. Assume that $0\leq x \leq \pi/2$.

Attempt at solution: Right now we are on the Bisection Method in Numerical Methods, so I believe this question must be related to a root finding question.
My attempt then is that the area is: $$\text{Area}= x\cos(x).$$ Taking the derivative using product rule then leads to: $$-x\sin(x)+\cos(x).$$ Setting this equal to 0 and re-arranging we get: $$cos(x)=x\sin(x)$$ Dividing by $\cos(x)$ by both sides. $$1=x\tan(x).$$ Then we have $$f(x)=\cot(x)-x.$$ Finding the roots of this: I get $x=0.86033.$

Concerns: I'm worried that I'm doing this incorrectly as if I type into wolfram alpha the roots of $-x\sin(x)+\cos(x)$, they give me answers denoting solution over the reals and numerical solutions. I'm not familiar with this distinction. Any help would be much appreciated. Thank you.


Your approach seems correct. Regarding Wolfram Alpha, “Solution over the reals” just means a solution in the real numbers $\mathbb{R}$ (as opposed to the complex numbers $\mathbb{C}$). The numerical solutions are just that—the numerical solutions (these are real numbers). In this case there are infinitely many, but you want the one in the range $[0,\pi/2]$, as you have already realized.

  • $\begingroup$ I see thank you! $\endgroup$ – kemb Jun 22 '18 at 2:19

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