## Maxima and Minima using calculus

Problem:

The vertices of an $(n+1)$-gon lie  on the sides of a regular n-gon and divide its perimeter into  parts of equal length. How should one construct the $(n+1)-$ gon so that its area is :

(a) maximum

(b) minimum

Hint only:

[One of the golden rule of solving problems in math/physics is to draw diagrams, as had benn emphasized by the maverick American physics Nobel Laureate, Richard Feynman. He expounded this technique even in software development. So, in the present problem, first draw several diagrams.]

There exists a side $B_{1}B_{2}$ of the $(n+1)$ -gon that lies entirely on a side $A_{1}A_{2}$ of the n-gon. Let $b=B_{1}B_{2}$ and $b=A_{1}A_{2}$. Show that $b=\frac{n}{n+1}a$. Then, for $x=A_{1}B_{1}$, we have $0 \leq x \leq \frac{n}{n+1}$ and the area S of the $(n+1)$ -gon is given by

$S(x)=\frac{\sin{\phi}}{2}\Sigma_{i=1}^{n}(\frac{i-1}{n+1}a+x)(\frac{n-i+1}{n+1}a-x)$

where $\phi=\angle{A_{1}A_{2}A_{3}}$. Thus, $S(x)$ is a quadratic function of x. Show that $S(x)$ is a minimal when $x=0$ or $x=\frac{a}{n+1}$ and $S(x)$ is maximal when $x=\frac{a}{2(n+1)}$.

Let me know if you have any trouble when you attempt it,

-Nalin Pithwa

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