Answer
The required dimensions are : $x=22 \ ft $ and $y=11 \ ft$
Work Step by Step
Let us consider that $x$ and $y$ be the length and width.
Given: $xy=242 \ sq. \ ft ~~~~(a)$
So, as per the given condition we can write as:
$C(x)=4(2y)+2(2x)\\ C(x)=8y+4x ~~~~~(b)$
Also, $xy=242 \implies y=\dfrac{242}{x}$
Equation (b) gives: $C(x)=\dfrac{1936}{x}+4x$
Foe minimum cost of fence we must have $\dfrac{dC(x)}{dx} =0$
So, $C'(x)=\dfrac{-1936}{x^2}+4=0 \implies x=22 \ ft $ and $y=\dfrac{242}{22}=11 \ ft$
Therefore, the required dimensions are : $x=22 \ ft $ and $y=11 \ ft$
