Let ABCD be the total page and A’B’C’D’ be the area in which the matter is printed.
Let ‘l’ and ‘b’ be the length and breadth of the total page.
We know that area of the rectangle is l × b
From the problem,
⇒ lb = 150cm2 …… (1)
⇒ PQ + RS = 3cm
⇒ WX + YZ = 2cm
Let ‘l’’ and ‘b’’ be the length and breadth of the area in which the matter is printed,
From the figure,
⇒ l’ = (l - 2)cm
⇒ b’ = (b - 3)cm
⇒ Area of the printed matter (A) = (l - 2)(b - 3)cm2
From (1)
We need Area of the printed matter maximum and let us take A as the function of l
We know that for maxima and minima,
⇒ I2 = 100
⇒ l = 10cm
(since length is an positive quantity)
Differentiating A again,
We get the area of the printed matter maximum for l = 10cm
Let’s find the corresponding breadth using eq (1)
⇒ 10b = 150
⇒ b = 15cm
∴ The dimensions required for the printed area to be maximum is l = 10cm and b = 15cm.