# Finding rectangle of maximum area (optimization/calculus)

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.

## 1 Answer

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.

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