Question

A rectangle has perimeter $26\text{inch}$, and the length and width are both whole numbers. What is the greatest possible area this rectangle may have in square $\text{inch}$?

Answer 1

Finding the required area:

Given: Perimeter $=26\text{in}$

Let the length and width of the rectangle be $\mathrm{l}\mathrm{cm}$ and $\mathrm{w}\mathrm{cm}$, respectively.

So, $2\left(\mathrm{l}+\mathrm{w}\right)=26$

$\Rightarrow$ $\mathrm{l}+\mathrm{w}=13$

Thus possible pairs of length and width are $\left(1,12\right),\left(2,11\right),\left(3,10\right),\left(4,9\right),\left(5,8\right),\left(6,7\right)$

Therefore, the greatest area of the rectangle is possible only with the pair $\left(6,7\right)$ of length and width

So, area of rectangle $=6\times 7$

$=42{\text{in}}^2$

Hence, the greatest possible area this rectangle may have is $42{\text{in}}^2$.