The top and bottom margins of a poster 66 cm each, and the side margins are 44 cm each. If the area of the printed material on the poster is fixed at 384384 square centimeters, find the dimensions of the poster of smallest area.

24cm by 36cm

Step-by-step explanation:

You may have written the values twice in the above question.

Let 'a' represent width of the poster and 'b' represent the height.

'A' represent the area of the poster to be minimized.

ab= 384cm²

b= 384/x

total height including top and bottom margins of a poster 6 cm each

b+12

total width including  the side margins 4 cm:

a+8

Therefore, Area of total poster would be defined as

A= (b+12)(a+8)

Substituting 'b' from above in equation on Area, we have

A=(a+8)(384/x + 12)

A= 384+12a+ (3072/a)  + 96

A= 12x + 3072/x + 480

A'= 12 - 3072/x²    (consider it A' for now)

considering A' =0

0= 12 - 3072/x²

3072/x² = 12

x²= 3071/12

x²=256

x=+-16

we'll ignore the negative root

therefore, x-16

Since A''= 2(3072)/x³ will be positive for x>0, A is concave up and x=16 is a minimum.

The total value of corresponding b will be,

b= 384/16 = 24cm

and the total width of the poster will be

x+8= 16+8= > 24cm

and the total height will be y+12

24+12=> 36cm

thus, the dimensions of the poster of smallest area is 24cm by 36cm

to find the perimeter of the rectangle this is the expression:

x-9+x-9+3x-4+3x-4=138

if we simplify it:

8x-26=138

remember, 138 is what it all adds up to

the length is 11.5cm

the width is 57.5cm

the area is 661.25cm^2

the answers might not be correct use at own risk

the answer is two. hope this !