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Answer 1 · 2017-09-28T09:04:10

(a) $(30x)/(x-4)+2x$ ; (b) $x>4$ and (c) $11.75$ inches

As width of page is $x$ and height of page is $y$ and

top and bottom margins are $1"$ and margins on each side are $2"$

the dimensions of print area are $x-4$ and $y-2$

and as print area is $30in^2$ , we have $y-2=30/(x-4)$ and $y=30/(x-4)+2$

(a) Hence area of page $A$ in terms of $x$ is

$x(30/(x-4)+2)$ or $(30x)/(x-4)+2x$

(b) We ought to have $x>4$ so that margins of $2"$ on either side are there.

(c) The graph of area vs. $x$ appears as

graph{(30x)/(x-4)+2x [-5, 25, 43.32, 83.32]}

and area is minimum at around $x~=11$ or $12$ nches.

Using calculus given $A=(30x)/(x-4)+2x$ , we have

$(dA)/(dx)=30((x-4-x)/(x-4)^2)+2=(-120+2(x-4)^2)/(x-4)^2$

and this is zero when $(x-4)^2=60$ and $x=sqrt60+4~=11.75$