Question

Maximum area of rectangle whose two vertices lies on the $x-$axis & two on the curve $y=3-|x|,\forall |x|<3$

• A. $\frac{9}{8}\text{sq. units}$
• B. $\frac{9}{4}\text{sq. units}$
• C. $3\text{sq. units}$
• D. $\frac{9}{2}\text{sq. units}$

Answer 1

The correct option is D $\frac{9}{2}\text{sq. units}$

$y=3-|x|$
$\Rightarrow y=\{\begin{array}{cc}3+x,& x<0\\ 3,& x=0\\ 3-x,& x>0\end{array}$
$\Rightarrow$Area of rectangle
$A=2x(3-x)=2(3x-x^2)$
$\Rightarrow$ For Maximum Area $\frac{dA}{dx}=0\Rightarrow x=\frac{3}{2}$
$\frac{d^{2}A}{d{x}^2}=-4<0$
$\Rightarrow x=\frac{3}{2}$ for maximum
$\Rightarrow A=\frac{2\left(9\right)}{4}=\frac{9}{2}\text{sq. units}$