You are designing a rectangular poster to contain 100 in2 of printing with a 4-in margin at the top and bottom and a 1-in margin at each side. what overall dimensions will minimize the amount of paper used?

  28 inches high by 7 inches wide

Step-by-step explanation:

Let x represent the width of the poster with margins. Then the printable width is (x -2). The printable height will be 100/(x-2), so the overall poster height is ...

  height = 100/(x -2) +8 = (8x +84)/(x -2)

The poster's overall area is the product of its width and height, so is ...

  A = x(8x +84)/(x -2)

The derivative of this with respect to x is ...

  A' = ((16x +84)(x -2) -(8x^2 +84x)(1))/(x -2)^2

This is zero when the numerator is zero, so ...

  8x^2 -32x -168 = 0

  x^2 -4x -21 = 0 . . . . . . divide by 8

  (x +3)(x -7) = 0 . . . . . . . factor

The values of x that make these factors be zero are -3 and +7. The height corresponding to a width of 7 is ...

  height = 100/(7 -2) +8 = 28

The amount of paper is minimized when the poster is 7 inches wide by 28 inches tall.

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Comment on the problem and solution

You will notice that the poster is 4 times as high as it is wide. It is no accident that this ratio is the ratio of the vertical margin to the horizontal margin. That is, the fraction of the poster devoted to margin is the same in each direction. This is the generic solution to this sort of problem.

Knowing that the margins have a ratio of 4:1 tells you the printable area will have a ratio of 4:1, hence is equivalent to 4 squares, each with an area of 100/4 = 25 square inches. That means the printable area is √25 = 5 inches wide by 4×5 = 20 inches high, so the overall poster area is 28 inches high by 7 inches wide. This arithmetic can be all mental and does not involve derivatives.

care to share were the trapezoid shown is ?