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.
