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A rectangle with one side lying along the x-axis is to be inscribed in the closed region of the plane bounded by the lines $y = 0, y = 3 x a n d y = 30 - 2 x .$ If M is the largest area of such a rectangle, then value of $\frac{2 M}{27}$ is

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Detailed Solution

$A = (x_{2} - x_{1}) y$

$H e x y = 3 x_{1} & y = 30 - 2 x_{2}$ $\Rightarrow x_{1} = \frac{y}{3} & x_{2} = \frac{30 - y}{2}$

$\begin{array}{rcl} A & = & (\frac{30 - y}{2} - \frac{y}{3}) y \\ = & 15 y - \frac{5 y^{2}}{6} \end{array}$

Differentiate and equate to zero

$A^{1} (y) = 0$

$15 - \frac{10 y}{6} = 0 \Rightarrow y = 9$

Therefore, the maximum area is $A = 135 - \frac{135}{2} = \frac{135}{2}$

Hence, $\frac{2}{27} (\frac{135}{2}) = \frac{135}{27} = 5$

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