The top and bottom margins of a poster are 6 cm and the side margins are each 8 cm. if the area of printed material on the poster is fixed at 384 square centimeters, find the dimensions of the poster with the smallest area. \

Ans : x = width of the printed material y = length of the printed material a = area of printed material on the poster = 384 cm^2 fixed A = area of poster we are trying to find A, which is x * y First off, let's start with w. We already know that the side margins are each 6 cm, so the width for the poster should be y + 12. In the same way, we know that the length of the poster should be x + 8. Now we can use the already given formula for area of the printed material: xy = 384 y = 384/x A = (x + 8) * (y + 12) A = (x + 8) * (384/x + 12) A = 384 + 12x + 3072/x + 96 Now we will find the optimal minimal area by setting the derivative of the area equal to 0: A' = 12 - 3072/x^2 12 - 3072/x^2 = 0 12 = 3072/x^2 12x^2 = 3072 x^2 = 256 x = 16 (it only makes sense that the result is positive so -16 is not a choice) Going back to the original equation, we have: xy = 384 16y = 384 y = 384/16 = 24 So we know that the dimensions of the printed material inside of the poster are 16 and 24, which are x and y, respectively. We were already given that the side margins corresponding to the width are measured to be 4 cm, and that the side margins corresponding to the length are measured to be 6 cm. Thus, we know that the width of the entire poster is 16 + 4 * 2 (for the two margins) = 24, and that the length of the entire poster is 24 + 6 * 2 = 36. The dimensions of the poster are 24 cm and 36 cm, the width and the length, respectively.

20y5 times 2 is 1010 times -2 is -20, plus y