Answer:
The dimensions of the page are
3.46 ft by 5.20 ft
Step-by-step explanation:
Let
x---> the length of the sheet of paper in feet
y ---> the width of the sheet of paper in feet
[tex]A=xy[/tex]
[tex]A=18\ ft^2[/tex]
so
[tex]18=xy[/tex]
[tex]y=\frac{18}{x}[/tex] -----> equation A
Remember that
[tex]1\ ft=12\ in[/tex]
Convert the margins into feet
[tex]9\ in=9\12=0.75\ ft[/tex]
[tex]6\ in=6\12=0.50\ ft[/tex]
so
we know that
The area of the largest printed area is given by
[tex]A=(y-0.75-0.75)(x-0.50-0.50)[/tex]
[tex]A=(y-1.50)(x-1)[/tex]
[tex]A=xy-y-1.50x+1.50[/tex]
substitute equation A in the above expression
[tex]A=x(\frac{18}{x})-\frac{18}{x}-1.50x+1.50\\[/tex]
[tex]A=18-\frac{18}{x}-1.50x+1.50[/tex]
[tex]A=19.50-\frac{18}{x}-1.50x[/tex]
Now we have an output (A) in terms of only one variable (x),
so
we differentiate:
[tex]\frac{dA}{dx}=\frac{18}{x^2}-1.50[/tex]
equate to zero
[tex]\frac{18}{x^2}=1.50[/tex]
[tex]x^2=12\\x=3.46\ ft[/tex]
Find the value of y
[tex]18=(3.46)y\\y=5.20\ ft[/tex]
therefore
The dimensions of the page are
3.46 ft by 5.20 ft
The required dimensions are,
[tex]x+18=3\sqrt{3}+18\\ y+12=2\sqrt{3}+12[/tex]
The formula of the area of the rectangle is [tex]A=l \times b[/tex]
Let [tex]A[/tex] be the area of the paper then,
[tex]A=(x+18)(y+12)...(1)[/tex]
And the printed area is [tex]xy=18...(2)[/tex]
Now, from the equation (1) and (2) we get,
[tex]A=(x+18)(\frac{18}{x}+12)\\ A=234+12x+\frac{324}{x} ..(3)[/tex]
Now, differentiating equation (3)
[tex]\frac{dA}{dx}=12-\frac{324}{x^2} \\\frac{dA}{dx}=0\\12-\frac{324}{x^2} =0\\x=3\sqrt{3}[/tex]
Substituting the obtained value of [tex]x[/tex] into the equation (2)
[tex]x+18=3\sqrt{3}+18\\ y+12=2\sqrt{3}+12[/tex]
Learn more about the topic area of the rectangle:
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