find the dimension

Question

the area of the page of a book is 40 square inches, the area of the printed part is 21 7/8 square inches. the margin at the top and sides is 3/4 inch wide, the margin at the bottom is 1 inch wide. find the dimension of the page.

Answer 1

Hi Emily,

 

You have two unknowns here.  Thus, you will need two equations.

 

In this case, it might be easier to follow if you let L and W represent that rectangular area of the text.  Imagine that you have two rectangles.  You have one rectangle that represents the whole page, and then you have a smaller rectangle inside that represents the area of the print.

 

If L is length of the printed text, then the length of the page must be L + 0.75 + 1.0 in = L + 1.75 in

If W is the width of the printed text, then width of the page must be W + 0.75 + 0.75 in = W + 1.5 in

 

So now you can set up your two equations.  You have one equation for the area of the printed text, recalling that the area of a rectangle is L*W.  You also have an equation for the area of the paper which is the longer length and width found above.

 

Don't want to work the whole problem!  Hope this helps.

Answer 2

page is x by y inches.

xy = 40; y = 40/x

21 7/8 = (x-3/2)(y-3/4-1)

21 7/8 = (x-3/2)(40/x–7/4)

21 7/8 = (1/2)(2x-3)(1/(4x))(160–7x)

Multiply both sides by 8x:

x(168 + 7) = (2x–3)(160–7x)

175x = -14x^2+320x+21x-480

14x^2 - (341 - 175)x + 480 = 0

14x^2 – 166x + 480 = 0

7x^2 – 83x + 240 = 0

a(x-h)^2+k = ax^2–2ahx+ah^2+k

h = 83/14

k = 240 – 7*(83/14)^2 = 240 – 83*83/(2*14)

k = (240*2*14 – 83*83)/(2*14) = -169

7(x–83/14)^2 = 169 = 13^2

| x–83/14 | = 13/√(7)

x = 83/14 ± 13/√(7)

x = 3.28282011750684 or 8.57432273963602

y = 40/x = 12.18464569127177 or 4.6650914847296

Two possibilities: 3.3 by 12.2 or 8.6 by 4.7