Question

The area of a rectangle will be maximum for the given perimeter when the rectangle is a

• A. Parallelogram
• B. Trapezium
• C. Square
• D. None of these

Answer 1

The correct option is C

Square

Explanation for Correct Answer:

Let, $ABCD$ is a rectangle with sides $x$and $y$.

The perimeter of a given rectangle is

$$\begin{gathered} p=x+y+x+y \\ =2x+2y \\ \Rightarrow 2y=p-2x \\ \Rightarrow y=\frac{p-2x}{2}\to \left(i\right) \end{gathered}$$

The area of a given rectangle is

$$\begin{gathered} A=xy \\ =x\left(\frac{p-2x}{2}\right) \\ =\frac{1}{2}\left(px-x^2\right) \end{gathered}$$

Differentiate the area with respect to $x$and equate it to zero, We get

$$\begin{gathered} \frac{dA}{dx}=\frac{1}{2}\left(p-2\times 2x\right) \\ =\frac{1}{2}\left(p-4x\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{1}{2}\left(p-4x\right)=0 \\ \Rightarrow p-4x=0 \\ \Rightarrow x=\frac{p}{4} \end{gathered}$$

Putting, $x=\frac{p}{4}$ in equation $\left(i\right)$ we get,

$$\begin{gathered} y=\frac{p-2\times {\displaystyle \frac{p}{4}}}{2} \\ =\frac{{\displaystyle \frac{p}{2}}}{2} \\ =\frac{p}{4} \end{gathered}$$

As we see that $x=\frac{p}{4}$ and $y=\frac{p}{4}$. Therefore the sides are equal which is possible only in the case of a square.

Therefore the area of a rectangle will be maximum for a given perimeter when the rectangle is a square.

Hence, the correct answer is option (C)