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

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