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

domain:

x =/= 0 x > 0 cannot have negative width

(o,oo) open interval

differentiate:

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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