You are designing a rectangular poster to contain 98 in2 of printing with a 4 inch margin at the top and botto?
Define the total area as A= (x +4 ) (y + 8) so that the print area is xy and y = 98/x A = (X + 4 ) . (98/X + 8) = 98 + 384/X + 8X + 32 = 8X + 384/X+ 130 dA/dX = 0 - 192/X^2 +8 = 0 => 192/X^2 = 8 => 8X^2 = 192 => X^2 = 24 => X= 4.9 and y= 98/4.9 = 20. Total paper required including margin = 4.9 + 4 and 20 + 8 = 8.9 x 28 in
The top and bottom margins of a rectangular poster?
The top and bottom margins of a rectangular poster are each 2 inches long, and the side margins are each 1/2 inch long. If the area of the printed material on the poster is fixed at 64 square inches, find the dimensions of the poster with the smallest area A. Also draw a picture of the situation. Please give details on how this would be solved. Thank you
You are designing a rectangular poster to contain 36in^2 of printing with a 4 inch margin?
Area(printed) = x * y = 36 => y = 36/x Area(total) = (x + 2)(y + 8) = xy + 8x + 2y + 16 = x(36/x) + 8x + 72/x + 16 A = 8x + 72/x + 52 A' = 8 - 72/x^2 = 0 x^2 - 9 = 0 x = 3 y = 12 Edit: Well, I gave you the minimized x & y , just plug-in (x + 2)(y + 8) , and what do you get? A = (3 + 2)(12 + 8) = 5 * 20 Looks better now?
Designing a poster?
printed area = 50 = xy paper size, P(x) = (x + 2*2)(y + 2*4) =(x + 4)(y + 8) = xy + 4y + 8x + 32 = 8x + 4(50/x) + (50 + 32) = 8x + 200/x + 82 P'(x) = 8 - 200/x^2 = 0 x = √(200/8) = 5 (for x > 0) answer : paper dimension = 9 x 18