Question

Show that of all the rectangles with a given perimeter, the one with the greatest area is a square.

Answer 1

Hint: For solving this question first we will assume the length and breadth of the rectangles in terms of two variables $x,y$ and then we will express the area of that rectangle as a function $f\left( x \right)$ of one variable $x$ . Then, we will use the condition of maxima that ${f}'\left( x \right)=0$ and ${f}''\left( x \right)<0$ to find the value of $x$ and then we will prove that for maximum area $x=y$ to prove the desired result.

Complete step-by-step answer:

Given: All rectangles whose perimeter is constant.
Now, let the length of the rectangle is $x$ units and breadth of the rectangle is $y$ units. And let the constant perimeter is $c$ units. Then,
$\begin{aligned} & 2x+2y=c \\\ & \Rightarrow x+y=\dfrac{c}{2} \\\ & \Rightarrow y=\dfrac{c}{2}-x............\left( 1 \right) \\\ \end{aligned}$
Now, the area of such a rectangle is equal to the product of length and breadth. Let $a$ represents the area of the rectangle. Then,
$a=xy$
Now, substituting $y=\dfrac{c}{2}-x$ from equation (1) in the above equation. Then,
$\begin{aligned} & a=xy=x\left( \dfrac{c}{2}-x \right) \\\ & \Rightarrow a=\dfrac{xc}{2}-{{x}^{2}} \\\ \end{aligned}$
Now, in the above equation, we have expressed the area in terms of one variable $x$ . So, we can write the area as a function of $x$ . Then,
$a=f\left( x \right)=\dfrac{xc}{2}-{{x}^{2}}...........\left( 2 \right)$
Before we proceed we should know how the derivative of the function $f\left( x \right)$ helps us in determining the maximum and minimum value of the function. If the maximum and minimum value of $f\left( x \right)$ occurs at $x={{x}_{1}}$ and $x={{x}_{2}}$ respectively. Then,
1. ${f}'\left( {{x}_{1}} \right)={{\left[ \dfrac{d\left( f\left( x \right) \right)}{dx} \right]}_{x={{x}_{1}}}}=0$ and ${f}''\left( {{x}_{1}} \right)={{\left[ \dfrac{d\left( {f}'\left( x \right) \right)}{dx} \right]}_{x={{x}_{1}}}}<0$ .
2. ${f}'\left( {{x}_{2}} \right)={{\left[ \dfrac{d\left( f\left( x \right) \right)}{dx} \right]}_{x={{x}_{2}}}}=0$ and ${f}''\left( {{x}_{2}} \right)={{\left[ \dfrac{d\left( {f}'\left( x \right) \right)}{dx} \right]}_{x={{x}_{2}}}}>0$ .
Now, in this question, we have to find the value of $x$ for which $f\left( x \right)=\dfrac{xc}{2}-{{x}^{2}}$ will be maximum. Then,
$\begin{aligned} & f\left( x \right)=\dfrac{xc}{2}-{{x}^{2}} \\\ & \Rightarrow {f}'\left( x \right)=\dfrac{c}{2}-2x=0 \\\ & \Rightarrow x=\dfrac{c}{4} \\\ \end{aligned}$
Now, as we have got $x=\dfrac{c}{4}$ for which the first derivative of $f\left( x \right)=\dfrac{xc}{2}-{{x}^{2}}$ is zero. So, we can check whether the second derivative is positive or negative for $x=\dfrac{c}{4}$ . Then,
$\begin{aligned} & f\left( x \right)=\dfrac{xc}{2}-{{x}^{2}} \\\ & \Rightarrow {f}'\left( x \right)=\dfrac{c}{2}-2x \\\ & \Rightarrow {f}''\left( x \right)=-2 \\\ & \Rightarrow {f}''\left( x \right)<0 \\\ \end{aligned}$
Now, as the second derivative is negative so, at $x=\dfrac{c}{4}$ the function $f\left( x \right)=\dfrac{xc}{2}-{{x}^{2}}$ will be maximum.
Now, we have got the value of $x$ for the maximum area so, from equation (1) we can find the value of $y$ . Then,
$\begin{aligned} & y=\dfrac{c}{2}-x \\\ & \Rightarrow y=\dfrac{c}{2}-\dfrac{c}{4} \\\ \end{aligned}$
$\Rightarrow y=\dfrac{c}{4}$
Thus, for the area to be the maximum value of $x=y=\dfrac{c}{4}$ . Which means the length and breadth of the rectangle should be equal for the maximum area. So, it will be a square.
Thus, for all the rectangles whose perimeter is constant, then the rectangle having maximum area will be a square.
Hence, proved.

Note: Here, the students should first understand what we have to prove then proceed in the right directions with proper assumptions. Moreover, we should apply the concept of maxima and minima accurately and don’t confuse the value of the first derivative and second derivative for each case. Then, equations should be solved stepwise to prove the result.