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## Homework Statement

The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. If the area of printed material on the poster is fixed at 384 cm

^{2}, find the dimensions of the poster with the smallest area.

## Homework Equations

Area of a rectangle is length times width

## The Attempt at a Solution

So I assumed that the 384 cm

^{2}formed a constraint like so:

[tex] (a-6)(b-4)=384[/tex]

and then using this equation I solved for a to be able to substitute it into the area formula:

[tex] ab-4a-6b+24=384 [/tex]

[tex] ab-4a=360+6b [/tex]

[tex] a(b-4)=360+6b [/tex]

[tex] a= \frac {360+6b}{b-4} [/tex]

so then i can substitute a in A=ab so that A will be a function of b:

[tex] A=ab [/tex]

[tex] A=\frac{360+6b}{b-4} b [/tex]

[tex] A(b)=\frac{360b+6b^{2}}{b-4} [/tex]

ok, so then i take the derivative of this A(b) now which turns out to be:

[tex] A'(b)=\frac {6b^{2}-48b-1440}{(b-4)^{2}} [/tex]

then i take the top, complete the square and then set it equal to zero:

[tex] 6(b-4)^{2}-1536 = 0 [/tex]

[tex] 6(b-4)^{2} = 1536 [/tex]

[tex] (b-4)^{2} = 256 [/tex]

[tex] b-4 = \sqrt{256}= \pm 16 [/tex]

[tex] b= \pm 16 +4 [/tex]

I chose to leave out the negative square root as it makes no sense when we are talking about lenth, so with b=20, take that back and plug it into the a=b equation:

[tex] a = \frac {360+6(20)} {20-4} = 30 [/tex]

and I get a = 20 and b=30 and if you take off the margins and multiply them at 16*24 you get 384 cm

^{2}, but alas, this is not correct. Where did I go wrong at?