Question

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

• A. $9$
• B. $\frac{9}{4}$
• C. $3$
• D. $\frac{9}{2}$

Answer 1

Answer: (D)

$\frac{9}{2}$

Let the base of the rectangle be denoted by the points on the x axis
$(x,0)$ and $(-x,0)$.
Then the rest of the two vertices will be $(-x,3-x),(x,3-x)$.
Hence length of the base of the rectangle$=2x.$
Height of the rectangle$=3-x.$
Hence area
$A=2x(3-x)$
$=6x-2x^2$
$\frac{dA}{dx}=6-4x$
$=0$
$x=\frac{3}{2}$.
Then
$Are{a}_{max}$
$=2x(3-x)$
$=2.\frac{3}{2}(3-\frac{3}{2})$
$=\frac{9}{2}$ sq units.