All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(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 $latex = 40(\frac{10}{9}) - 18(\frac{10}{9})^{2} = \frac{200}{9}$

KS Comments

Students should bare in mind to check the second order to substantiate that its a maximum area.

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