Explanation

(d)
$\because$ Perimeter of a sector$= p$
Let $AOB$ be the sector with radius $r$
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If angle of the sector be $\theta$ radius, then area of sector,
$A = \frac{1}{2}r^{2}\theta$ …(i)
And length of arc, $s = r\theta$
$\Rightarrow \ \ \ \ \theta = \frac{s}{r}$
$\therefore$ perimeter of the sector
$p = r + s + r = 2r + s$ …(ii)
On subtracting $\theta = \frac{s}{r}$ in Eq. (i), we get
$A = \left( \frac{1}{2}r^{2} \right)\left( \frac{s}{r} \right) = \frac{1}{2}rs$
$\Rightarrow \ \ s = \frac{2A}{r}$
Now, on substituting the value of $s$ in Eq. (ii), we get
$p = 2r + \left( \frac{2A}{r} \right)\ \ \Rightarrow \ \ \ 2A = pr - 2r^{2}$
On differentiating w.r.t. $r$, we get
$2\frac{dA}{dr} = p - 4r$
For the maximum area, put
$2\frac{dA}{dr} = 0\ \Rightarrow \ \ p - 4r = 0\ \ \Rightarrow \ \ r = \frac{p}{4}$