Question

The total area of a page is 150 cm². The combined width of the margin at the top and bottom is 3 cm and the side 2 cm. What must be the dimensions of the page in order that the area of the printed matter may be maximum?

Answer 1

\[\text { Let x andybe the length and breadth of the rectangular page, respectively. Then,}\]

\[\text { Area of the page } = 150\]

\[ \Rightarrow xy = 150\]

\[ \Rightarrow y = \frac{150}{x} . . . \left( 1 \right)\]

\[\text { Area of the printed matter }=\left( x - 3 \right)\left( y - 2 \right)\]

\[ \Rightarrow A = xy - 2x - 3y + 6\]

\[ \Rightarrow A = 150 - 2x - \frac{450}{x} + 6\]

\[ \Rightarrow \frac{dA}{dx} = - 2 + \frac{450}{x^2}\]

\[\text { For maximum or minimum values of A, we must have }\]

\[\frac{dA}{dx} = 0\]

\[ \Rightarrow - 2 + \frac{450}{x^2} = 0\]

\[ \Rightarrow 2 x^2 = 450\]

\[ \Rightarrow x = 15\]

\[\text { Substituting the value of x in  }\left( 1 \right), \text { we get }\]

\[y = 10\]

\[\text { Now }, \]

\[\frac{d^2 A}{d x^2} = \frac{- 900}{x^3}\]

\[ \Rightarrow \frac{d^2 A}{d x^2} = \frac{- 900}{\left( 15 \right)^3}\]

\[ \Rightarrow \frac{d^2 A}{d x^2} = \frac{- 900}{3375} < 0\]

\[\text { So, area of the printed matter is maximum when  x = 15 and y = 10 } . \]