Question

The perimeter of a given rectangle is x, its area will maximum when its side are:

• A. $\frac{x}{2},\frac{x}{2}$
• B. $\frac{x}{3},\frac{x}{3}$
• C. $\frac{x}{6},\frac{x}{3}$
• D. $\frac{x}{4},\frac{x}{4}$

Answer 1

Answer: (D)

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’
$\therefore A^{\prime }=\frac{dA}{db}=\frac{x}{2}-2b$
Put $A^{\prime }=0\Rightarrow \frac{x}{2}-2b=0\Rightarrow b=\frac{x}{4}$
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 $\frac{x}{4},\frac{x}{4}$