y is proportional to $x^n$. x and y are the length and breadth of a rectangle with a fixed area.

1. Express the fixed area condition as a proportional relationship

For a rectangle with fixed area,
$xy=\text{constant}.$
Hence
$y=\frac{\text{constant}}{x},$
so $y$ is inversely proportional to $x$. Therefore $y\propto x^{-1}$.

2. Identify the exponent in $y\propto x^n$

Comparing $y\propto x^{-1}$ with $y\propto x^n$ gives
$n=-1.$

1. Find the new breadth as a fraction of the original breadth when the length is doubled

Since the area is fixed, $xy=\text{constant}$, so $y\propto \dfrac{1}{x}$.

When $x$ is doubled, $y$ is multiplied by $\dfrac{1}{2}$.

So the new breadth is $\dfrac{1}{2}$ of the original breadth.

2. Calculate the percentage reduction in breadth

Reduction in breadth
$=1-\frac{1}{2}=\frac{1}{2}.$

Percentage reduction
$=\frac{1}{2}\times 100\%=50\%.$