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Application of Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives1 Mark
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).
(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
$
\begin{aligned}
& x+x+x+l+l=12 \\
\Rightarrow & 3 x+2 l=12 \Rightarrow l=6-\frac{3}{2} x ...(i)
\end{aligned}
$
Let $A m ^2$ be the area of the window corresponding to the above dimensions, then
$
\begin{aligned}
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))
\end{aligned}
$
Now, $\frac{d A}{d x}=6-3 x+\frac{\sqrt{3}}{2} x$
$
\begin{array}{l}
\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
\end{array}
$
Now, $\frac{d^2 A}{d x^2}=-3+\frac{\sqrt{3}}{2}<0$
$\therefore \quad A$ 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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