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Answer 1 · 2017-09-28T09:04:10

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

Explanation:

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

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

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

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

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

$x (\frac{30}{x - 4} + 2)$ $x(30/(x-4)+2)$ or $\frac{30 x}{x - 4} + 2 x$ $(30x)/(x-4)+2x$

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

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

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

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

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

$\frac{d A}{d x} = 30 (\frac{x - 4 - x}{{(x - 4)}^{2}}) + 2 = \frac{- 120 + 2 {(x - 4)}^{2}}{{(x - 4)}^{2}}$ $(dA)/(dx)=30((x-4-x)/(x-4)^2)+2=(-120+2(x-4)^2)/(x-4)^2$

and this is zero when ${(x - 4)}^{2} = 60$ $(x-4)^2=60$ and $x = \sqrt{60} + 4 ≅ 11.75$ $x=sqrt60+4~=11.75$

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