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A rectangular page is to have a printed area of 62 square inches. If the border is to be 1 inch wide on top and bottom and only 1/2 inch wide on each side find the dimensions of the page that will use the least amount of paper

Can someone explain how to do this?

I started with:

$$A = (x + 2)(y + 1) $$

Then I isolate y and come up with my new equation:

$$A = (x+2)\left(\frac{62}{x + 2}{-1}\right)$$

Then I think my next step is to create my derivative, but wouldn't it come out to -1?

Anyways, I would appreciate if someone could give me a nudge in the right direction.


How does this look for a derivative?

$$A = \left(\frac{x^2-124}{x^2}\right)$$

Then to solve: $$ {x} = 11.1 $$

$$ y = 98 / 11.1 $$

Does that seem about right?

If not, the only thing I would have left is setting it to 0 and solving.

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Hint: We have $xy=62$. Don't know where your $98/(x+2)$ came from, it is not right. So we want to minimize $(x+2)(\frac{62}{x}+1)$. – André Nicolas May 3 '12 at 21:47
$A_{printed} = (h_{total} - 2)(w_{total} - 1)$ – Joel Cornett May 3 '12 at 21:48
@AndréNicolas Sorry that was a type, I fixed it :) Why is it $62$ over $x$? Rather than $62$ over $x + 2$ – Math_Phase May 3 '12 at 21:58
Because the printed area is $62$ square inches, and (without explicitly saying so) you let $x$ and $y$ be the height and width of the printed area. (That's how you got the expression $(x+2)(y+1)$.) You should always say so, first of all to tell the reader, and almost as importantly, to remind yourself. Your expression is not fully fixed. It is $\frac{62}{x}$ and plus $1$. – André Nicolas May 3 '12 at 22:01
@AndréNicolas My derivative looks slightly off, am I doing something wrong when using the product rule? – Math_Phase May 3 '12 at 22:36
up vote 2 down vote accepted

Hint: How did you get the term $\left(\frac {98}{x+2}-1\right)$? You should have $62=xy$ to give the desired printable area, so $A=(x+2)(\frac{62}x+1)$. Then, you are right, you should take $\frac {dA}{dx}$ and set it to $0$ to find $x$.

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Your response was first and lead me in the right direction. Thank you! – Math_Phase May 3 '12 at 23:06

If the dimensions of the printed area are $x$ and $y$, where $y$ is the dimension with the $1/2$ inch borders (the "width"), then the printed area is $$\tag{1}62= x y.$$ You want to minimize the area of the entire page, which is $$\tag{2}A=(x+2)(y+1).$$ We want $A$ expressed in terms of one variable only; so solve $(1)$ for $y$ $$\tag{3} y={62\over x } $$ and substitute into $(2)$, giving $$\tag{4} A(x) = (x+2)\cdot\textstyle\bigl( {62\over x }+1\bigr) . $$ Now you want to minimize $A(x)$ over $x\in(0,\infty)$. Do this using the normal derivative analysis (remember to examine what happens when $x$ is close to $0$ and when $x$ is big).

Once you've found the value of $x$ that minimizes $(4)$, remember to state the answer to the question explicitly; for example "the dimensions of the paper are $x+2$ inches top to bottom and $y+1$ inches wide" (you can use $(3)$ to find the value of $y$ once you have $x$.

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Thank you for the reply. Once I register, I'll be sure to come back and upvote. – Math_Phase May 3 '12 at 23:06

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