MCQ
Assertion $(A) :$ A window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is $12 m$, then length $1.782 m$ and breadth $2.812 m$ of the rectangle will produce the largest area of the window.
Reason $( R )$ : For maximum or minimum, $f^{\prime}(x)=0$.
  • Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A).$
  • B
    Both $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of $(A).$
  • C
    $(A)$ is true but $(R)$ is false.
  • D
    $(A)$ is false but $(R)$ is true.

Answer

Correct option: A.
Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A).$
Let each side of the equilateral triangle be $x m$ and $l m$ be the height (length) of the rectangular part of the window, then
$x+x+x+l+l=12$
$\Rightarrow 3 x+2 l=12$
$\Rightarrow l=6-\frac{3}{2} x ...(i)$
Let $A m ^2$ be the area of the window corresponding to the above dimensions, then
$A =l x+\frac{\sqrt{3}}{4} x^2 ; x>0, l>0$
$ =\left(6-\frac{3}{2} x\right) x+\frac{\sqrt{3}}{4} x^2 ($ Using $(i))$
Now, $\frac{d A}{d x}=6-3 x+\frac{\sqrt{3}}{2} x$
$\frac{d A}{d x}=0$
$\Rightarrow 6-3 x+\frac{\sqrt{3}}{2} x=0$
$\Rightarrow 12-6 x+\sqrt{3} x=0$
$\Rightarrow x=\frac{12}{6-\sqrt{3}} \approx 2.812$
Now, $\frac{d^2 A}{d x^2}=-3+\frac{\sqrt{3}}{2}<0$
$\therefore A$ has local maxima at $x=2.812$
For $x=2.812, l=6-\frac{3}{2}(2.812)=1.782$
$\therefore$ Height of rectangular part $=1.782 m$ and breadth $=2.812 m$

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