Optimization practice problem question 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.
EDIT 
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.
 A: 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 $x$.
A: 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 $x$.
