SOLUTION: 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 a

Algebra ->  Test -> SOLUTION: 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 a      Log On


   

Question 982771: 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
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
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.
:
Let L = the length of the printed material
Let w = the width of the printed material
We know L*W = 64
L = 64%2Fw
:
The 2" top and bottom adds 4" to length
the 1/2 sides add 1" to width
therefore the area of the poster will be
A = (L+4)*(w+1)
FOIL
A = Lw + L + 4w + 4
We know Lw = 64, therefore
A = 64 + L + 4w + 4
A = L + 4w + 68
replace L with 64%2Fw
A = 64%2Fw + 4w + 68
:
plot this equation
+graph%28+300%2C+200%2C+-6%2C+20%2C+-20%2C+200%2C+%2864%2Fx%29%2B4x%2B68%29+
minimum area occurs when w = 4" is the width of the text
then
64/4 = 16" is the length of the text
:
"find the dimensions of the poster with the smallest area A."
16+4 = 20" the length of the poster, (2" margin top and bottom)
and
4 + 1 = 5" the width of the poster, (1/2" margin on each side)