# Re: Precalculus Question

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Posted by Mary on September 28, 2002 at 22:01:47:

In Reply to: Re: Precalculus Question posted by Brad Paul on September 28, 2002 at 18:35:38:

: : Hi,

: : I'm can't figure out how to approach this question"

: : Suppose we want to make a poster containing 50 square inches of printed matter surrounded by a 3" border at the top and bottom, and a 2" border along each side. How does the total area of the poster depend on the width of the printed matter?.

: : Any clues, anyone? Please?

: Let A be the area of the printed matter, and w the width of the
: printed matter. If the printed matter is in the shape of a rectangle
: we can find the height of the printed matter via h=A/w. Let b be the
: 3 inch border at the top and bottom and c be the 2 inch width of the
: border along each side. Now with these variable lets write an equation
: for the total area of the poster. Lets call it Atotal

:
: Atotal=(2 c+w)(2 b+A/w)

: Notice this equation is just a height time width equation. The width
: is 2 time the border width because there is one on each side and the
: the width of the printed matter. The height is 2 times the borders
: plus height written in terms of the Area and width of the printed
: matter.

It wouldn't be:
A = (6+w)(4+w) and therefore A = 24 + w(10+w) since we know the printed matter is a square and the complete poster is a rectangle? Also, how would i show this as a function of w? Would it be expressed as A(w) = 24 + w(10+w) since i need to show how the total area is dependent on the width of the printed material.....Thanks!

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