The total area of a page

Solution:

Let $x$ and $y$ be the length and breadth of the rectangular page, respectively. Then,

Area of the page $=150$

$\Rightarrow x y=150$

$\Rightarrow y=\frac{150}{x}$                             ......(1)

Area of the printed matter $=(x-3)(y-2)$

$\Rightarrow A=x y-2 x-3 y+6$

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

$\Rightarrow \frac{d A}{d x}=-2+\frac{450}{x^{2}}$

For maximum or minimum values of $A$, we must have

$\frac{d A}{d x}=0$

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

$\Rightarrow 2 x^{2}=450$

$\Rightarrow x=15$

Substituting the value of $x$ in $(1)$, we get

$y=10$

Now,

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

$\Rightarrow \frac{d^{2} A}{d x^{2}}=\frac{-900}{(15)^{3}}$

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

So, area of the printed matter is maximum when $x=15$ and $y=10$.