A rectangular page is to contain 16 square inches of print. The margins on each side are 1 inch. Find the dimensions of the page such that the least amount of paper is used?

Question

(Answer should be small and large)

Isaiah Allen · Accepted Answer

For the least usage of paper, the reqd. dimns. of the page, are,

#"Length="(l+2)=6# inch, and, the #"Width="(16/l+2)=6# inch.

Let #l# inch be the length of printed rectangular region of the page. Since, the Area of the printed rectangular region has to be #16# sq.in., we find that, the width of the prited portion must be #16/l# inch. Now, the margin of #1# inch has been left on both sides, so, the length of the page must be #(l+2)# inch, and, smilarly, the width, # (16/l+2)# inch. These give us, the Area of the page #(l+2)(16/l+2)=16+2l+32/l+4,# #or, 20+2(l+16/l),# which, being a fun. of #l,# we write, it as, #A(l)=20+2(l+16/l).........(1)# To find the least amt. of paper, we need to minimise #A(l).# Knowing that, for #A_(min), A'(l)=0, and, A''(l) >0.# #"From "(1), A'(l)=0:. 2{1-16/l^2}=0:.l^2=16:.l=+-4#. #A''(l)=2{0-16*(-2)l^-3}=64/l^3 rArr A''(+4)=1 >0.# #:. l=+4" gives "A_(min).# Thus, for the least usage of paper, the reqd. dimns. of the page, are, #"Length="(l+2)=6# inch, and, the #"Width="(16/l+2)=6# inch. Enjoy Maths.!

Christopher Agresta · Answer

undefined Let $ x $ be the width of the printed area and $ y $ be the length of the printed area. Since there are margins of 1 inch on each side, the total width of the page is $ x + 2 $ inches and the total length is $ y + 2 $ inches.

Given that the total area of the printed page is 16 square inches, we have the equation:

$$ xy = 16 $$

We want to minimize the amount of paper used, which is the total area of the page. So, we need to minimize $ (x + 2)(y + 2) $.

Now, we'll solve for $ y $ from the first equation and substitute it into the expression for the total area:

$$ y = \frac{16}{x} $$

$$ \text{Total area} = (x + 2)\left(\frac{16}{x} + 2\right) $$

To find the minimum, we'll take the derivative with respect to $ x $, set it to zero, and solve for $ x $:

$$ \frac{d}{dx}\left[(x + 2)\left(\frac{16}{x} + 2\right)\right] = 0 $$

$$ \frac{d}{dx}\left[16 + 32x^{-1} + 2x + 4\right] = 0 $$

$$ -32x^{-2} + 2 = 0 $$

$$ 32x^{-2} = 2 $$

$$ x^{-2} = \frac{1}{16} $$

$$ x = 4 $$

Now that we have $ x = 4 $, we can find $ y $:

$$ y = \frac{16}{x} = \frac{16}{4} = 4 $$

So, the dimensions of the page such that the least amount of paper is used are $ \boxed{4 \times 4} $ inches.