The least length of the thread required to construct a rectangle of area 256 $c m^{2}$ is

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Solution

The correct option is B 64
area $= a b$
$256 = a b$
$2 (a + b) = p$
$2 (a + \frac{256}{a}) = p$
use AM & GM
$\frac{a + \frac{256}{a}}{2} \geq (a \times \frac{256}{a})^{\frac{1}{2}}$
$a + \frac{256}{a} \geq 32$
$p \geq 64$

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The least length of the thread required to construct a rectangle of area 256 cm2 is

The correct option is B 64area =ab256=ab2(a+b)=p2(a+256a)=puse AM & GMa+256a2≥(a×256a)12a+256a≥32p≥64

The least length of the thread required to construct a rectangle of area 256 $c m^{2}$ is

The correct option is B 64
area $= a b$
$256 = a b$
$2 (a + b) = p$
$2 (a + \frac{256}{a}) = p$
use AM & GM
$\frac{a + \frac{256}{a}}{2} \geq (a \times \frac{256}{a})^{\frac{1}{2}}$
$a + \frac{256}{a} \geq 32$
$p \geq 64$