Posted by Daniel on Sunday, November 20, 2011 at 11:06pm.
The top and bottom margins of a paster are each 6 cm and the side margins are each 4cm. If the area of printed material on the poster is fixed at 384cm^2, find the dimensinos of the poster with the smallest area.
I would like to know if my work is correct.
x = width of print
y = height of print
w = width of print
h = height of print
w = x + 8
h = y + 12
adding both sides of magins
A = wh
A = (x+8)(y + 12) xy = 384 y = 384/x
A = (x+8)(384/x + 12)
A = (x+8)([384 + 12x]/x)
A = (12x^2 + 480x + 3072)/x
x =/= 0 x > 0 cannot have negative width
(o,oo) open interval
A'(x) = (24x + 480)/-x^2
and this is where I get stuck, x is not able to be 0 yet I keep getting -20 to make the function = 0. Is the function actually a closed interval? I am actually having a hard time figuring out if something is an open or closed interval
- calculus - Steve, Monday, November 21, 2011 at 5:21am
Your A' is wrong. I expect the quotient rule got messy. Just make it a sum of powers of x:
A = 12x + 3072/x + 480
A' = 12 - 3072/x^2
A' = 0 when 12 = 3072/x^2
or, x = 16
so, y = 24
The printing is 16x24 and the poster is 24x36
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