Question

The top and bottom margins of a poster 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, find the dimensions of the poster of smallest area. (list the smallest dimension first)

Answer 1

Answer:

p = 25,20 cm

h = 34,32 cm

A(min) = 864,86 cm²

Step-by-step explanation:

Let call p and h dimensions of a poster, ( length , and height respectively) and x and y  dimensions of the printed area of the poster then

p = x + 8        and      h = y + 12

Printed area = A(p) = 384 cm²     and   A(p) = x*y     ⇒ y = A(p)/x

y = 384 / x

Poster area  =  A(t) = ( x + 8 ) * ( y + 12 ) ⇒ A(t) =  ( x + 8 ) * [( 384/x ) + 12 ]

A(t)  =  384 + 12x + 3072/x + 96      A(t) = 480 + 3072/x + 12x

A(t)  =  [480x + 12x² + 3072 ] / x

A´(t) =  [(480 + 24x )* x  - (480x + 12x² + 3072]/x²

A´(t) =  0         [(480 + 24x )* x  - 480x - 12x² - 3072] =0

480x + 24x² -480x -12x² - 3072 = 0

12x² = 3072         x² = 296

x = 17,20 cm       and      y =  384/17,20      y = 22,32 cm

Notice  if you substitu the value of x = 17,20 in A(t) ; A(t) >0 so we have a minimun at that point

Then dimensions of the poster

p = 17,20 + 8  = 25,20 cm

h = 22.32 + 12 =34,32 cm

A(min) = 25,20 *34.32  

A(min) = 864,86 cm²