Optimizing Page Dimensions for Minimum Paper Usage

In summary, the margins of a book will be 3 cm at the top and bottom and 2 cm on the sides, with a total printed area of 81 cm2. To minimize the amount of paper used, the page dimensions should be x = 6 cm and y = 16 cm.
  • #1
Fifty
35
0
"The printed area of a book will be 81 cm2. The margins at the top and bottom of the page will each be 3 cm deep. The margins at the sides of the page will each be 2 cm wide. What page dimensions will minimize the amount of paper?" (Calculus and Vectors, Peter Crippin et al., 158).

[I added the citation just in case!]

I first thought...
"the width of the entire page will be x and the length will be y. The width of the printed area then is x - 4 (excluding the side margins) and the length of the printed area is y - 6. So (x - 4)(y - 6) = 81 cm2.

The total area is A = xy. y, in terms of x, is y = [itex]\frac{1000}{x - 4}[/itex] + 6. So then I figured:

A(x) = x ( [itex]\frac{1000}{x - 4}[/itex] + 6 ) would be the area and I would take its derivative and set A'(x) = 0."

I always end up with 81x - 81x in the differentiated function and get 0 = -324. I've tried multiple approaches (including setting x to the width of the printed page and then the total width being x + 4) and using different methods to get to the derivative, but this always happens.

Thanks for the help!

EDIT: Whoops! Thanks, chet!

unsimplified:
A'(x) =[itex]\frac{81(x-4) - 81x}{(x - 4)}[/itex]

Sorry, apparently formatting the fraction won't work if I include the exponent formatting as well, but the (x - 4) in the denominator is supposed to be squared.
 
Last edited:
Physics news on Phys.org
  • #2
Fifty said:
"The printed area of a book will be 81 cm2. The margins at the top and bottom of the page will each be 3 cm deep. The margins at the sides of the page will each be 2 cm wide. What page dimensions will minimize the amount of paper?" (Calculus and Vectors, Peter Crippin et al., 158).

[I added the citation just in case!]

I first thought...
"the width of the entire page will be x and the length will be y. The width of the printed area then is x - 4 (excluding the side margins) and the length of the printed area is y - 6. So (x - 4)(y - 6) = 81 cm2.

The total area is A = xy. y, in terms of x, is y = [itex]\frac{1000}{x - 4}[/itex] + 6. So then I figured:

A(x) = x ( [itex]\frac{1000}{x - 4}[/itex] + 6 ) would be the area and I would take its derivative and set A'(x) = 0."

I always end up with 81x - 81x in the differentiated function and get 0 = -324. I've tried multiple approaches (including setting x to the width of the printed page and then the total width being x + 4) and using different methods to get to the derivative, but this always happens.

Thanks for the help!
It's hard to help unless you show us what you got for the derivative.

Chet
 
  • #3
Fifty said:
"The printed area of a book will be 81 cm2. The margins at the top and bottom of the page will each be 3 cm deep. The margins at the sides of the page will each be 2 cm wide. What page dimensions will minimize the amount of paper?" (Calculus and Vectors, Peter Crippin et al., 158).

[I added the citation just in case!]

I first thought...
"the width of the entire page will be x and the length will be y. The width of the printed area then is x - 4 (excluding the side margins) and the length of the printed area is y - 6. So (x - 4)(y - 6) = 81 cm2.

The total area is A = xy. y, in terms of x, is y = [itex]\frac{1000}{x - 4}[/itex] + 6. So then I figured:

If [itex](x - 4)(y - 6) = 81[/itex] then surely [itex]y = \frac{81}{x - 4} + 6[/itex]?

A(x) = x ( [itex]\frac{1000}{x - 4}[/itex] + 6 ) would be the area and I would take its derivative and set A'(x) = 0."

I always end up with 81x - 81x in the differentiated function and get 0 = -324. I've tried multiple approaches (including setting x to the width of the printed page and then the total width being x + 4) and using different methods to get to the derivative, but this always happens.

Thanks for the help!

EDIT: Whoops! Thanks, chet!

unsimplified:
A'(x) =[itex]\frac{81(x-4) - 81x}{(x - 4)^2}[/itex]

(LaTeX fixed)

I think you're missing a [itex]6(x-4)^2[/itex] from the numerator; you should have [itex]A'(x) = xy'(x) + y(x)[/itex].
 
  • #4
pasmith said:
If [itex](x - 4)(y - 6) = 81[/itex] then surely [itex]y = \frac{81}{x - 4} + 6[/itex]?



(LaTeX fixed)

I think you're missing a [itex]6(x-4)^2[/itex] from the numerator; you should have [itex]A'(x) = xy'(x) + y(x)[/itex].

I see what I did wrong; I didn't differentiate correctly. I got it now, thanks :D
 

1. What is calculus?

Calculus is a branch of mathematics that focuses on the study of change and motion. It is used to analyze and model continuous change, such as the motion of objects, rates of change, and slopes of curves.

2. What are the two main branches of calculus?

The two main branches of calculus are differential calculus and integral calculus. Differential calculus deals with the study of rates of change and slopes of curves, while integral calculus deals with the computation of areas and volumes under curves.

3. What are some real-life applications of calculus?

Calculus has many real-life applications, such as in physics (for calculating motion and forces), engineering (for designing structures and systems), economics (for analyzing supply and demand), and medicine (for modeling the growth of tumors).

4. What are some important concepts in introductory calculus?

Some important concepts in introductory calculus include limits, derivatives, and integrals. Limits are used to define the behavior of functions at certain points, derivatives measure the rate of change of a function, and integrals measure the area under a curve.

5. How can I improve my understanding and skills in calculus?

To improve your understanding and skills in calculus, it is important to practice solving problems regularly. Additionally, seeking help from a tutor or attending extra help sessions can also be beneficial. It is also important to have a good understanding of algebra and trigonometry, as they are the foundation for calculus concepts.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
459
  • Calculus and Beyond Homework Help
Replies
6
Views
949
  • Calculus and Beyond Homework Help
Replies
1
Views
3K
  • Calculus and Beyond Homework Help
Replies
6
Views
5K
  • Calculus and Beyond Homework Help
Replies
2
Views
5K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Math Proof Training and Practice
2
Replies
56
Views
7K
  • Math Proof Training and Practice
2
Replies
48
Views
9K
  • Calculus and Beyond Homework Help
Replies
3
Views
4K
Back
Top