Question

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 on the left and right are to be 1½ cm as shown below:

On the basis of the above information, answer the following questions:

1. Write the expression for the area of the visiting card in terms of x.
2. Obtain the dimensions of the card of minimum area.

Answer 1

(i) Given, x = length of printed area

y = breadth of printed area

Area of printed part = xy

or, 24 = xy

or, y = `24/x`       ...(i)

Now, area of visiting card = (x + 3) (y + 2)

or, A = (x + 3)(y + 2)

or, A = (x + 3)`(24/x + 2)`

or, A = `((x + 3)(24 + 2x))/x`

or, A = `(2x^2 + 30x + 72)/x`

or A = 2x + 30 + `72/x`

(ii) We have, Area, A = 2x + 30 + `72/x`

∴ `(dA)/dx = 2 + 0 -72/x^2`

and `(d^2A)/(dx^2) = 144/x^3`

for minimum area, put `(dA)/dx = 0`

i.e., `2-72/x^2 = 0`

or, x² = 36 ⇒ x = ± 6

At x = `6.(d^2A)/(dx^2)>0` Hence, minimum

Therefore, length of vising card = x + 3 = 6 + 3 = 9 cm

and breadth of visiting card = y + 2 = `24/x + 2`

= `24/6 + 2`

= 6 cm