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A metal wire 108 meters long is bent to form a rectangle. If the area of the rectangle
is maximum, then its dimensions are
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is maximum, then its dimensions are
Solution:
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Verified Answer
The correct answer is:
27 m, $27 \mathrm{~m}$
Let sides of rectangle be $x$ and $y$
Thus $2 x+2 y=108 \Rightarrow x+y=54 \Rightarrow y=54-x$
Now Area $=A=x y$
$\therefore \quad=x(54-x)=54 x-x^{2}$
Differentiating w.r.t. $\mathrm{x}$, we get
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=54 \times 1-2 \mathrm{x}$ and $\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=-2 < 0$
When $\frac{d A}{d x}=0$, we get $54-2 x=0 \Rightarrow x=27 \Rightarrow y=54-x=27$
Thus $2 x+2 y=108 \Rightarrow x+y=54 \Rightarrow y=54-x$
Now Area $=A=x y$
$\therefore \quad=x(54-x)=54 x-x^{2}$
Differentiating w.r.t. $\mathrm{x}$, we get
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=54 \times 1-2 \mathrm{x}$ and $\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}=-2 < 0$
When $\frac{d A}{d x}=0$, we get $54-2 x=0 \Rightarrow x=27 \Rightarrow y=54-x=27$
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