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(i)
Perimeter $= 3x + 4x + 2y + 5x = 20$

$\Rightarrow y = 10 -6x$

Area, $S = 4xy + \frac{1}{2}(3x)(4x) = 40 x - 18x^{2}$

(ii)
$\frac{dS}{dx} = 40 - 36x = 0$

$\Rightarrow x = \frac{10}{9}$

Since $latex \frac{d^{2}S}{dx^{2}} = -36 < 0$, S is maximum when $latex x = \frac{10}{9}$ Required Area $latex = 40(\frac{10}{9}) - 18(\frac{10}{9})^{2} = \frac{200}{9}$