To find the dimensions of the rectangular poster with the given specifications, we need to account for the margins and the area of the printed matter. Let’s denote the width of the printed matter as w inches and the height as h inches.
According to the problem, the printed area is 50 square inches:
w * h = 50
Next, we need to include the margins in our total dimensions. The poster will have side margins of 2 inches on each side, resulting in an overall width of:
W = w + 2 + 2 = w + 4
The top and bottom margins are 4 inches each, so the overall height will be:
H = h + 4 + 4 = h + 8
Now we want to minimize the area of the poster:
A = W * H = (w + 4)(h + 8)
Substituting the expression for h from the printed area equation, we get:
h = 50 / w
Thus, the area expression becomes:
A = (w + 4) * (50/w + 8)
Expanding this:
A = (w + 4)(50/w + 8) = 50 + 8w + 200/w + 32
A = 82 + 8w + 200/w
To find the minimum area, we take the derivative of A with respect to w and set it equal to zero:
A’ = 8 – 200/w^2
Setting the derivative to zero:
0 = 8 – 200/w^2
200/w^2 = 8
w^2 = 25
w = 5
Now substituting back to find h:
h = 50 / 5 = 10
Finally, we can determine the overall dimensions of the poster:
Width of poster = 5 + 4 = 9 inches
Height of poster = 10 + 8 = 18 inches
Therefore, the dimensions of the poster of least area that meets these specifications are:
Width: 9 inches, Height: 18 inches