The top and bottom margins of a poster are 2 cm and the side margins are each 4 cm. if the area of printed material on the poster is fixed at 388 square centimeters, find the dimensions of the poster with the smallest area.

The poster will have minimum area when the margins are the same fraction of total width in each direction. That is, the poster will be twice as wide as it is tall.

That condition requires the width of the printed area of the poster to be
  width = √(2*388 cm²) = 27.86 cm
so the total width will be
  27.86 cm + 2×4 cm = 35.86 cm
The height will be half that,
  height = width/2 = 17.93 cm

The poster with the smallest area will be 36.86 cm wide by 17.93 cm high.


Let A represent the printable area of the poster, and let mx and my represent the total margin width in the x (horizontal) and y (vertical) directions, respectively. Then the area of the poster with margins (P) as a function of poster width w is ...
  P = w*(A/(w -mx) +my)
We want to find w so that the derivative dP/dw is zero.
  dP/dw = (A/(w -mx) +my) -w*(A/(w -mx)^2) = 0
  -mx*A/(w -mx)^2 +my = 0
  (w -mx)^2 = A(mx/my)

Then the overall poster width is ...
  w = mx +√(A*mx/my) . . . . . . as described above.

Not so sure about this one sorry.