What are the dimensions of the page such that the least amount of paper is used?

Question

A rectangular page is to contain 36 square inches of print.
The margins on each side are to be 1 1/2 inches.

Answer 1

9" x 9" is the best size.

If there are 36 square inches of print, then the possible dimensions of the page (length x width) are:

1 x 36
2 x 18
3 x 12
4 x 9
6 x 6
9 x 4
12 x 3
18 x 2
36 x 1

Since there has to be a 1.5" margin on each side, we add 3" to the length (1.5" on the top and 1.5" on the bottom) and 3" to the width (1.5" on the left and 1.5" on the right). Do this for all the possible dimensions.

So our new dimesions are:

4 x 39 = (1 + 3) x (36 + 3)
5 x 21 = (2 + 3) x (18 + 3)
6 x 15 = ...
7 x 12 = ...
9 x 9 = ...
12 x 7 = ...
15 x 6 = ...
21 x 5 = (18 + 3) x (2 + 3)
39 x 4 = (36 + 3) x (1 + 3)

Now figure out the number of square inches for each possible dimension.

4 x 39 = 156
5 x 21 = 105
6 x 15 = 90
7 x 12 = 84
9 x 9 = 81
12 x 7 = 84
15 x 6 = 90
21 x 5 = 105
39 x 4 = 156

So you can see that the dimensions that give you the smallest area are 9" x 9". So that's the answer.

Answer 2

Usually I would just say a square, but since there are margins there may be something else going on here with regards to the paper used to print area ratio. When that ratio is at its lowest then we know that we have the correct answer.

Paper used to print area ratio: (Paper used) / (Print Area)

We know that Print Area is going to be 36, so thats easy enough to substitute in: (Paper used) / 36sq. in.

We also know that the Paper Used is a function of the total area, so its Width x Height: (W)*(H) / 36sq. in.

So now we have an equation - W*H / 36 - However there are some restrictions on the Width and Height of the paper, the width and height must be made in such a fasion that the inner "print area" is 36 square inches. We can do this by knowing that the "inner area" will have a width and height that are both 3 inches less than the outer width and height, respectively, because there is 1 1/2 inches on each side. Therefore (W-3)*(H-3) = 36. Next to simplify the ratio eqation we can subtitute one variable for the other by solving for it in the variable requirements equation:

(W-3)*(H-3) = 36

(W-3) = 36/(H-3) "divide both sides by (H-3)

W = 36/(H-3) + 3 "add 3 to both sides, W is solved for and can be substituted into the ratio

(W)*(H) / 36sq. in. "Ratio

[36/(H-3) +3] * (H) / 36 "W Substituted in

[3H + (36*H)/(H-3)] / 36 "simlified

H/12 + H/(H-3) "Further simplified

Ratio = H/12 + H/(H-3) (We also know that H must be greater than 3 for an inner print area to be formed, however there is no upper limit to H)

Now since we know a definite equation for the ratio, and we want to find the dimensions that minimize that ratio, we can do this by taking the derivative of the ratio function and setting it to zero.

Function: F(H) = (1/12)*H + H/(H-3)

Derivative: F'(H) = 1/12 + (-3)/(H-3)^2

Set derivative equal to 0:

0 = 1/12 + (-3)/(H-3)^2

Multiply both sides by (H-3)^2:

0 = (H-3)^2/12 + (-3)

Mulitply both sides by 12:

0 = (H-3)^2 - 36

Simplify:

0 = H^2 -6H -27

Factor:

(H-9)(H+3) = 0

Solutions:

H = 9 , -3

Since we know that H > 3, we can throw out the -3 solution
So H must equal 9 to have the ratio of paper to print area be most efficient. This also means that Width will equal 9. This seems to be the same answer everyone else got.

Answer 3

Least amount of paper used means least area of paper. Area of paper is least when the paper is square. 36 sq. inches translate into 6x6 area. Add 11/2 inches on all sides. the result is
9 "x9".

Answer 4

x = top = backside y = facets (x - 6)(y - 2) = 28 y = 2(x + 8)/(x - 6) A(x) = 2x(x+8)/(x-6) A'(x) = 2(x^2 - 12x - 40 8)/(x - 6)^2 A'(x) = 0 if x^2 - 12x - 40 8 = 0 x = 2?21 + 6 ? 15.17 in y = 2?21/3 + 2 ? 5.06 in

Answer 5

9 x 9 inch.

anything else you want to know ?