A rectangular visiting card is to contain 24 sq.cm. of printed matter. The margins at the top and bottom of the card are to be 1 cm and the margins…

Let $\displaystyle \mathrm{A}(\mathrm{x})$ be the area of the visiting card then,

$$\text { As } \mathrm{x} y=$\displaystyle 24$, \mathrm{A}(\mathrm{x})=(\mathrm{x}+$\displaystyle 3$)(y+$\displaystyle 2$)=$\displaystyle 2$ \mathrm{x}+$\displaystyle 3$ y+\mathrm{x} y+$\displaystyle 6$=$\displaystyle 2$ \mathrm{x}+\frac{72}{\mathrm{x}}+$\displaystyle 30$

$$(ii) $\displaystyle \mathrm{A}^{\prime}(\mathrm{x})=2-\frac{72}{\mathrm{x}^{2}}$ and $\displaystyle \mathrm{A}^{\prime \prime}(\mathrm{x})=\frac{144}{\mathrm{x}^{3}}$,

solving $\displaystyle \mathrm{A}^{\prime}(\mathrm{x})=0 \Rightarrow \mathrm{x}=6$ is the critical point.

$$\mathrm{A}^{\prime \prime}($\displaystyle 6$)=\frac{144}{6}>$\displaystyle 0$, \therefore \text { Area of the card is minimum at } \mathrm{x}=$\displaystyle 6$, y=$\displaystyle 4$

$$The dimension of the card with minimum area is Length $\displaystyle =9 \mathrm{~cm}$, Breadth $\displaystyle =6 \mathrm{~cm}$