Find the dimensions of the rectangle of largest area having fixed perimeter 100.
Let length of rectangle $= \ell$
Let breadth of rectangle $= b$
Then, perimeter of rectangle $=2\left(\ell + b\right)$
Given: Perimeter of rectangle $= 100$
$\implies$ $2\left(\ell + b\right) = 100$
i.e. $\ell + b = 50$
$\implies$ $b = 50 - \ell$ $\;\; \cdots$ (1)
Area of rectangle $=A=\ell \; b$
i.e. $A = \ell \left(50 - \ell\right)$ $\; \;$ [in view of equation (1)]
i.e. $A = 50 \ell - \ell^2$ $\;\; \cdots$ (2)
We have from equation (2)
$\dfrac{dA}{d\ell} = 50 - 2\ell$ $\;\; \cdots$ (3)
For maximum area, $\dfrac{dA}{d\ell} = 0$
$\therefore$ From equation (3), $\dfrac{dA}{d\ell} = 50 - 2 \ell = 0$
$\implies$ $\ell = 25$
Now, from equation (3), $\dfrac{d^2 A}{d\ell^2} = -2 < 0$
$\implies$ $\ell = 25$ gives the length of a rectangle with maximum area.
Substituting the value of $\ell$ in equation (1) gives
$b= 50-25 = 25$
$\therefore$ Length of rectangle $= 25$ units; breadth of rectangle $= 25$ units