Man, who is right,I've got 2c answers that makes sense but the back of the book says?

saddy

New member
A printed page will have 2-in margins of white spaceon the sides and 1-in margins of white space on the top and bottom. the area of the printed portion is 32 in^2. Determine the dimensions of the page so that the least amount of paper is used. The answer(i only need the function, i don't need to solve this from the back of the book) a(x)= 40 +4x +(64/ x)....I don't even know how they got this...How can I solve other ones like this tooo...?answer1:A = area of paper
h = height of printed area
w = width of printed area
h*w = 32

A = (w + 2*2)(h + 2*1)
substitute w = 32/h
A = (32/h + 4)(h + 2)
A = 32 + 64/h + 4h + 8
A = 40 + 64/h + 4h


or answer2:
This is hard without a diagram, but imagine two rectangles, one inside the other. The inner one is your printed portion, and the outer one is the page. The printed portion is 32 sq in. So let's designate the width of the page x.
A = width*height, A = 32, and width = x.
32 = x*height
height = 32/x
So therefore, for our inner rectangle, width = x, height = 32/x.

The outer rectangle is the inner rectangle + margins.
width(outer) = x + two side margins = x + 4
height (outer) = 32/x + top margin + bottom margin = 32/x + 2

Since you're looking for the area function, just multiply the width and height of the outer rectangle:
A(x) = (x + 4)(32/x + 2)
Now apply FOIL to distribute these
A(x) = 32 + 2x + 128/x + 8
Combine like terms
A(x) = 40 + 2x + 128/x
are they both correct?
 
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