The top and bottom margins of a poster are 8 cm and the side margins are each 8 cm. If the area of printed material on the poster is fixed at 382 square centimeters, find the dimensions of the poster with the smallest area?
see related questions below
To find the dimensions of the poster with the smallest area, we can start by assuming that the width of the printed material on the poster is x cm.
Given that the top and bottom margins are 8 cm and the side margins are also 8 cm, we can calculate the total width and height of the poster.
Total width = Printed material width + Left margin + Right margin
Total width = x + 8 + 8 = x + 16 cm
Total height = Printed material height + Top margin + Bottom margin
Total height = 382 / x + 8 + 8 = 382 / x + 16 cm
Now, to find the area of the poster, we multiply the width and height:
Area = Total width × Total height
= (x + 16) × (382 / x + 16)
= 382 + 16x + 6144 / x + 16
To find the dimensions of the poster with the smallest area, we need to find the minimum value of the area. We can do this by finding the derivative of the area function, setting it equal to zero, and solving for x.
Let's differentiate the area function with respect to x:
d(Area)/dx = 16 - 6144 / (x + 16)^2
Setting the derivative equal to zero:
16 - 6144 / (x + 16)^2 = 0
Rearranging the equation:
6144 / (x + 16)^2 = 16
Multiplying both sides by (x + 16)^2:
6144 = 16(x + 16)^2
Taking the square root of both sides:
√(6144) = 4(x + 16)
Simplifying:
|x + 16| = √(6144) / 4
x + 16 = √(6144) / 4 or x + 16 = -√(6144) / 4
Solving for x:
x = (√(6144) / 4) - 16 or x = (-√(6144) / 4) - 16
Now that we have the value of x, we can substitute it into the formulas for the total width and height to find the dimensions of the poster with the smallest area.