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(C) Let the height of the rectangle be $y$. The top two vertices of the rectangle lie on the lines $y = 3x$ and $y = 30 - 2x$.For the left vertex on $

Vedclass · Accepted Answer

$\frac{135}{2}$

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(C) Let the height of the rectangle be $y$. The top two vertices of the rectangle lie on the lines $y = 3x$ and $y = 30 - 2x$.For the left vertex on $y = 3x$,we have $x_1 = \frac{y}{3}$.For the right vertex on $y = 30 - 2x$,we have $2x_2 = 30 - y$,so $x_2 = \frac{30 - y}{2} = 15 - \frac{y}{2}$.The width of the rectangle is $w = x_2 - x_1 = 15 - \frac{y}{2} - \frac{y}{3} = 15 - \frac{5y}{6}$.The area of the rectangle is $A(y) = w \cdot y = \left( 15 - \frac{5y}{6} \right) y = 15y - \frac{5y^2}{6}$.To find the maximum area,we take the derivative with respect to $y$ and set it to zero:$A'(y) = 15 - \frac{10y}{6} = 15 - \frac{5y}{3} = 0$.Solving for $y$,we get $\frac{5y}{3} = 15$,which implies $y = 9$.The second derivative $A''(y) = -\frac{5}{3} Substituting $y = 9$ into the area formula:$A_{max} = 15(9) - \frac{5(9^2)}{6} = 135 - \frac{5 \cdot 81}{6} = 135 - \frac{5 \cdot 27}{2} = 135 - \frac{135}{2} = \frac{135}{2}$.Thus,the largest area is $\frac{135}{2}$.

$A$ rectangle with one side lying along the x-axis is to be inscribed in the closed region of the $xy$ plane bounded by the lines $y = 0$,$y = 3x$,and $y = 30 - 2x$. The largest area of such a rectangle is

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