# Thread: Solving Dimensions given Area

1. ## Solving Dimensions given Area

A page contains 32 square inches of print. The margins at the top and the bottom of the page are 2 inches. The margins on each side are only 1 inch. Find the dimensions of the page so that the least paper is used.

I was taught to write two equations, one for perimeter and one for area.
I know that the equation for area is $\displaystyle A=(x-2)(y-4)=32$. Since I want the least paper used, I need to find the minimum of perimeter. But I don't know what I should put for the equation.

A rectangle is bounded by the x-axis and the semicircle $\displaystyle y=\sqrt{16-x^2}$. What length and width should the rectangle have so that its area is at a maximum?

I'm not sure how to show you the image, but its a graph with a semicircle on top of the x-axis in the middle of the y-axis with radius reaching out to (-4,0) and (4,0) from the origin. There's a rectangle within this semicircle.
So the length is split by the y-axis forming an x on each side. The width, y, stays the same. The area equation I wrote was A=2xy=max which is correct. Since I'm given y, I replace it in the equation giving me: $\displaystyle A=2x(\sqrt{16-x^2})$, and take the derivative to get $\displaystyle A'=\frac{-2x^2}{\sqrt{16-x^2}}+2\sqrt{16-x^2}$. So I set it equal to zero, and get the values $\displaystyle x=-2\sqrt{2}$ and $\displaystyle x=2\sqrt{2}$. But my teacher tells me there are more x values than that.

2. For the first problem, it might help to draw a diagram. You are told that on a printed page the dimensions of PRINT (the words) is 32 square inches:

$\displaystyle 32=x\cdot y$

We are also told that this page that printed text lies on, has 2 inch margins on the top, and 1 inch margins on the side. Thus the dimensions of the page are:

$\displaystyle l=x+2; w=y+4$

$\displaystyle A_{entire page}=(x+2)(y+4)$

Now in order to make a clean equation for the area of the ENTIRE page (not the printed area), we need to create an equation in a single variable. Going back to the first equation:

$\displaystyle 32=x\cdot y \Longrightarrow \frac{32}{x}=y$

Now we can plug into our equation for the area of the entire page:

$\displaystyle A_{entire page}=(x+2)(\frac{32}{x}+4) \Longrightarrow \\* A_{entire page}=(x+2)(\frac{32+4x}{x}) \Longrightarrow \frac{4x^2+40x+64}{x}$

Now we need to take the derivative of our function for the area of the entire page:

$\displaystyle A'_{entire page}=\frac{(8x+40)x-(4x^2+40x+64)(1)}{x^2} \Longrightarrow$$\displaystyle A'_{entire page}=\frac{8x^2+40x-4x^2-40x-64}{x^2} \Longrightarrow$$\displaystyle A'_{entire page}=\frac{4x^2-64}{x^2}$

From here we set the equation equal to zero, to find the minimum:

$\displaystyle 0=\frac{4x^2-64}{x^2} \Longrightarrow 64=4x^2 \Longrightarrow \pm 4=x$

One of these solutions, is obviously not possible. From here it is simply a matter of plugging the value of X into our original equation to find the length equal to 6. Plugging X into the equation we used to eliminate Y allows us to find that dimensions of the printed work to be 8, and thus the dimensions of the entire pages width to be equal to 12.

3. Hello, DarkestEvil!

A page contains 32 inē of print.
The margins at the top and the bottom of the page are 2 inches.
The margins on each side are only 1 inch.
Find the dimensions of the page so that the least paper is used.

I was taught to write two equations, one for perimeter and one for area.
No, that is for a specific (and different) problem.
Code:
      : 1 : - x - : 1 : -
_ *---------------*  -
: |               |  :
2 |               |  :
: |               |  :
- |   *-------*   |  :
: |   |///////|   |  :
: |   |///////|   |  :
y |   |///////|y  | y+4
: |   |///////|   |  :
: |   |///////|   |  :
- |   *-------*   |  :
: |       x       |  :
2 |               |  :
: |               |  :
- *---------------*  -
: - -  x+2  - - :

Let: $\displaystyle x$ = width of printed area.
Let: $\displaystyle y$ = height of printed area.

We are told that: .$\displaystyle xy \,=\,32 \quad\Rightarrow\quad y \,=\, \frac{32}{x}\;\;{\color{blue}[1]}$

The width of the page is: .$\displaystyle x+2$
The height of the page is: .$\displaystyle y+4$

The area of the page is: .$\displaystyle A \:=\:(x+2)(y+4)$

Substitute [1]: .$\displaystyle A \;=\;(x+2)\left(\tfrac{32}{x} + 4\right)$

And we have: .$\displaystyle A \;=\;4x + 64x^{-1} + 40$

And that is the function we must minimize.