The adjacent sides of a rectangle with given perimeter as 100 cm and enclosing maximum area areA- 10 cm and 40 cmB- 20 cm and 30 cmC- 15 cm and 35 cmD- 25 cm and 25 cm

The correct option is D 25 cm and 25 cm2x+2y=100 ⇒ x +y =50 ....(i) Let area of rectangle is A; ∴ A =xy ⇒ y =A/x Put in (i), we have x+A/x=50 ⇒ A=50 x x^2 ⇒d ...

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Standard XII

Mathematics

Local Maxima

The adjacent ...

Question

The adjacent sides of a rectangle with given perimeter as 100 cm and enclosing maximum area are

10 cm and 40 cm

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20 cm and 30 cm

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25 cm and 25 cm

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15 cm and 35 cm

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Solution

The correct option is C 25 cm and 25 cm
$2 x + 2 y = 100 \Rightarrow x + y = 50 . . . . (i)$
Let area of rectangle is A; $∴ A = x y \Rightarrow y = \frac{A}{x}$
Put in (i), we have $x + \frac{A}{x} = 50 \Rightarrow A = 50 x - x^{2}$
$\Rightarrow \frac{d A}{d x} = 50 - 2 x$
For maximum area $\frac{d A}{d x} = 0$
$∴ 50 - 2 x = 0 \Rightarrow x = 25$ and y =25
Hence adjacent sides are 25 and 25 cm.

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The adjacent sides of a rectangle with given perimeter as 100 cm and enclosing maximum area are

The correct option is C 25 cm and 25 cm2x+2y=100⇒x+y=50....(i) Let area of rectangle is A; ∴A=xy⇒y=Ax Put in (i), we have x+Ax=50⇒A=50x−x2 ⇒dAdx=50−2x For maximum area dAdx=0 ∴50−2x=0⇒x=25 and y =25 Hence adjacent sides are 25 and 25 cm.