Optimizing Page Dimensions for Minimum Paper Usage

[Fifty]

"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 = \frac{1000}{x - 4} + 6. So then I figured:

A(x) = x ( \frac{1000}{x - 4} + 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) =\frac{81(x-4) - 81x}{(x - 4)}

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.

 

[Chestermiller]

"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 = \frac{1000}{x - 4} + 6. So then I figured:

A(x) = x ( \frac{1000}{x - 4} + 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

 

[pasmith]

"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 = \frac{1000}{x - 4} + 6. So then I figured:

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

A(x) = x ( \frac{1000}{x - 4} + 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) =\frac{81(x-4) - 81x}{(x - 4)^2}

(LaTeX fixed)

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

 

[Fifty]

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



(LaTeX fixed)

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

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