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Q.
The perimeter of a given rectangle is x, its area will maximum when its side are:
Application of Derivatives
Solution:
Given: Perimeter of rectangle = x
$\Rightarrow \, 2(\ell + b) = x$
$ \Rightarrow \, \ell = \frac{x}{2} - b$
where $\ell$ is length of rectangle and b is breadth of rectangle.
Now, Area = length $\times$ breadth
$ \Rightarrow A = \ell \times b = \left(\frac{x}{2} - b \right) \times b $
$\Rightarrow A = \frac{x}{2} b - b^{2}$
Differentiate both side w.r.t ‘b’
$ \therefore A' = \frac{dA}{db} = \frac{x}{2} -2b $
Put $A' = 0 \Rightarrow \frac{x}{2} - 2b = 0 \Rightarrow b = \frac{x}{4}$
and $ A" = - 2 <0$
$ \therefore $ Area will be maximum at point $b = \frac{x}{4} $
$\therefore \ell = \frac{x}{2} - \frac{x}{4} = \frac{x}{4}$
$ \therefore $ Area of rectangle will be maximum if its sides are $ \frac{x}{4} , \frac{x}{4} $