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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?
Solution
\[\text { Let x andybe the length and breadth of the rectangular page, respectively. Then,}\]
\[\text { Area of the page } = 150\]
\[ \Rightarrow xy = 150\]
\[ \Rightarrow y = \frac{150}{x} . . . \left( 1 \right)\]
\[\text { Area of the printed matter }=\left( x - 3 \right)\left( y - 2 \right)\]
\[ \Rightarrow A = xy - 2x - 3y + 6\]
\[ \Rightarrow A = 150 - 2x - \frac{450}{x} + 6\]
\[ \Rightarrow \frac{dA}{dx} = - 2 + \frac{450}{x^2}\]
\[\text { For maximum or minimum values of A, we must have }\]
\[\frac{dA}{dx} = 0\]
\[ \Rightarrow - 2 + \frac{450}{x^2} = 0\]
\[ \Rightarrow 2 x^2 = 450\]
\[ \Rightarrow x = 15\]
\[\text { Substituting the value of x in }\left( 1 \right), \text { we get }\]
\[y = 10\]
\[\text { Now }, \]
\[\frac{d^2 A}{d x^2} = \frac{- 900}{x^3}\]
\[ \Rightarrow \frac{d^2 A}{d x^2} = \frac{- 900}{\left( 15 \right)^3}\]
\[ \Rightarrow \frac{d^2 A}{d x^2} = \frac{- 900}{3375} < 0\]
\[\text { So, area of the printed matter is maximum when x = 15 and y = 10 } . \]
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