Question

Find the length and width of a rectangle with maximum area that has a perimeter of $9P$ units.

Answer 1

Step-1: Forming expression for perimeter:

Considering length of rectangle $=x$

And width of rectangle $=y$

Then perimeter of rectangle $2x+2y=9P...\left(i\right)\left[\therefore perimeter=9Pisgiven\right]$

$$\begin{gathered} \Rightarrow 2y=9P-2x \\ \Rightarrow y=\frac{9P-2x}{2}....\left(i\right) \end{gathered}$$

Step-2: Forming expression for area:

Since we know the area of rectangle $A=xy$

$$\begin{gathered} \Rightarrow A=x\left(\frac{9P-2x}{2}\right) \\ \Rightarrow A=\frac{9Px-2x^2}{2} \end{gathered}$$

Differentiating $A$ with respect

$\Rightarrow \frac{dA}{dx}=\frac{9P}{2}-2x$

For maximum area

$$\begin{gathered} \Rightarrow \frac{dA}{dx}=0 \\ \Rightarrow \frac{9P}{2}-2x=0 \\ \Rightarrow 2x=\frac{9P}{2} \\ \Rightarrow x=\frac{9P}{4} \end{gathered}$$

And

$\Rightarrow \frac{d^{2}A}{d{x}^2}=-2<0$

Therefore area will be maximum at $x=\frac{9P}{4}$

Step-3: Finding maximum area:

Putting value of $x$ in equation (ii)

$$\begin{gathered} \Rightarrow 2\left(\frac{9P}{4}\right)+2y=9P \\ \Rightarrow 2y=9P-\frac{9P}{2} \\ \Rightarrow y=\frac{9P}{4} \end{gathered}$$

Hence, the required maximum area $xy=\left(\frac{9P}{4}\right)\left(\frac{9P}{4}\right)=\frac{81P^2}{16}$ square unit.