An advertisement consistsof a rectangular printed region plus 1in margins on the sides and 2in margins at top and bottom. If the area of the printed region is to be 92in^(2), find the dimensions of the printed region and overall advertisement that minimize the total aread.
2007-01-24 23:19:07 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Let a = length of printed region
Let b = width of printed region
Area of printed region = ab = 92
Therefore, b = 92 / a
Total area of whole advertisement is: T = (a + 2)(b + 4)
Expanding gives: T = ab + 4a + 2b + 8
But ab = 92 and b = 92/a
Therefore, T = 92 + 4a + 184/a + 8
or, T = 4a + 184a^(-1) + 100
Take the derivative and set it equal to zero:
dT/da = 4 - 184/a^2 = 0
Therefore, a = sqrt(46) and so b = 2*sqrt(46).
These are the dimensions of the printed square.
The total area = [sqrt(46) + 2][2*sqrt(46) + 4]
= 100 + 8*sqrt(46)
2007-01-25 01:05:30 · answer #1 · answered by falzoon 7 · 0⤊ 0⤋
you have to be able to write down the equation that descibes the area the total area and the printed area would be two separate equations (I think).
Anyway, when you get the equations, you'll have to take a derivative of them and set that derivative to zero (to find where the slope is zero means that you can solve for a minimal point on the curve of the original equation.
Hint: A square is also a rectangle.
good luck. I'm not going to solve the whole thing for you. I hope you understood what I said up there !!!
tom
2007-01-25 07:32:08 · answer #2 · answered by a1tommyL 5 · 0⤊ 0⤋