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. (list the smallest dimension first).
Width=?
height=?

Step-by-step explanation:

When margins are involved, the smallest area will be the one that has its dimensions in the same proportion as the margins. If x is the "multiplier", the dimensions of the printed area are ...

  (4x)(6x) = 384 cm^2

  x^2 = 16 cm^2 . . . . . divide by 24

  x = 4 cm

The printed area is 4x by 6x, so is 16 cm by 24 cm. With the margins added, the smallest poster will be ...

  24 cm by 36 cm

Comment on margins

It should be obvious that if both side margins are 4 cm, then the width of the poster is 8 cm more than the printed width. Similarly, the 6 cm top and bottom margins make the height of the poster 12 cm more than the height of the printed area.

Alternate solution

Let w represent the width of the printed area. Then the printed height is 384/w, and the total poster area is ...

  A = (w+8)(384/w +12) = 384 +12w +3072/w +96

Differentiating with respect to w gives ...

  A' = 12 -3072/w^2

Setting this to zero and solving for w gives ...

  w = √(3072/12) = 16 . . . . matches above solution.

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Generic solution

If we let s and t represent the side and top margins, and we use "a" for the printed area, then the above equation becomes the symbolic equation ...

  A = (w +s)(a/w +t)

  A' = t - sa/w^2

For A' = 0, ...

  w = √(sa/t)

and the height is ...

  a/w = a/√(sa/t) = √(ta/s)

Then the ratio of width to height is ...

  w/(a/w) = w^2/a = (sa/t)/a

  width/height = s/t . . . . . . the premise we started with, above

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier