A publisher decides that each page in a new book must have an area of 73.125 in^2, a 0.75 in. margin at the top and at the bottom of each page, and a 0.5 in. margin on each of the sides. What should the outside dimensions of each page be so that the printed area is a maximum?
Thanks!
Thanks!
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73.125 = x * (73.125/x) where 73.125 is the size of the entire page. Let x be the height of the entire page.
The dimensions of the written area are therefore:
A = (x - 1.5) * ((73.125/x) - 1)..........Multiply that out (carefully)
A = 73.125 - x - (109.6875/x) + 1.5 = 74.625 - x - (109.6875 * x^(-1))
dA/dx = -1 + (109.6875 * x^(-2))....................... Set this equal to 0.
-1 + (109.6875 * x^(-2)) = 0
(109.6875 * x^(-2)) = 1
109.6875 = x^2
x = 10.47318 is the original height
The original width was (73.125/x) = (73.125/10.47318) = 6.98212
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So the entire height is 10.47318
The entire width is 6.98212
**************************************…
.
The dimensions of the written area are therefore:
A = (x - 1.5) * ((73.125/x) - 1)..........Multiply that out (carefully)
A = 73.125 - x - (109.6875/x) + 1.5 = 74.625 - x - (109.6875 * x^(-1))
dA/dx = -1 + (109.6875 * x^(-2))....................... Set this equal to 0.
-1 + (109.6875 * x^(-2)) = 0
(109.6875 * x^(-2)) = 1
109.6875 = x^2
x = 10.47318 is the original height
The original width was (73.125/x) = (73.125/10.47318) = 6.98212
**************************************…
So the entire height is 10.47318
The entire width is 6.98212
**************************************…
.