Question

The top and bottom margins of a poster are 6 cm each, and the side margins are 4 cm each. If the area of the printed material on the poster is fixed at 384 square centimeters, the dimensions of the poster of the smallest area would be _____.

Answer 1

Answer:

The dimensions of the poster of the smallest area would be 24×36 cm.

Step-by-step explanation:

Let x and y be the width and height respectively of the printed area, which has fixed area.

If the area of the printed material on the poster is fixed at 384 square centimeters.

i.e. $xy=384\ cm^2$

$\Rightarrow y=\frac{384}{x}$

The top and bottom margins of a poster are 6 cm each is y+1.

The total width of the poster including 4 cm at sides is x+8.

The area of the total poster is $A=(x+8)(y+12)$

Substitute the value of y,

$A=(x+8)(\frac{384}{x}+12)$

$A=384+12x+\frac{3072}{x}+96$

$A=12x+\frac{3072}{x}+480$

Derivate w.r.t x,

$A'=12-\frac{3072}{x^2}$

Put A'=0,

$12-\frac{3072}{x^2}=0$

$\frac{3072}{x^2}=12$

$x^2=\frac{3072}{12}$

$x^2=256$

$x=16$

Derivate again w.r.t x,

$A''=\frac{2(3072)}{x^3}$ is positive for x>0,

A is concave up and x=16 is a minimum.

The corresponding y value is

$y=\frac{384}{16}$

$y=24$

The total poster width is x+8=16+8=24 cm

The total poster height is y+12=24+12=36 cm

Therefore, the dimensions of the poster of the smallest area would be 24×36 cm.