# The top and bottom margins of a poster are 6cm and the side margins are 4cm each. If the area of printed material on the poster is fixed at 384cm sq, find the dimension of the poster with the...

The top and bottom margins of a poster are 6cm and the side margins are 4cm each. If the area of printed material on the poster is fixed at 384cm sq, find the dimension of the poster with the smallest area.

### 2 Answers | Add Yours

Let's start with what we know. Most posters are rectangular so we will assume this one is as well. Let's let "L" be the length of the printed section and "W" be the width of the printed section. Therefore, the area of our printed section is L(W)=384. Once we know this, we can solve for one of the variables. Divide both sides by L and we will get:

W = 384/L

Let's hold onto that thought for a moment. If the top and bottom both have margns of 6 cm, then the expression we can use for the length of the overall poster is L + 12. Likewise, if our poster has right and left margins of 4 cm. each, then the expression we can use for the width of the overall poster is W + 8. Does this make sense?

To find the area of our overall poster, we will multiply the two expressions we just created for our poster length and width:

A = (L + 12)(W + 8)

So far, so good. Now, we can substitute the expression we developed earlier for W into this equation:

A = (L + 12)(384/L + 8)

Multiplying this out using a method to multiply binomials (i.e. "Man With The Big Nose," "FOIL," or Lattice Multiplication), we get:

384 + 8L + 4608/L + 96

We can combine this and write this out using L to the positive and negative powers:

8L + 4608L^(-1) + 480

To find the minimum, we need to take the derivative of this function. To do this, we see that L is raised to the first power, so we take what is in front of L (8) and multiply it by 1 and get 8. We would then subtract 1 from the exponent which would give us 0 so our answer for this part is just 8. We can eliminated the constant 480 because we cannot take the derivative of a constant and it does not impact the curve. So, the derivate of 4608L(^-1) would be -1(4608)^-2.

So, our derivative is:

-4608L^-2 + 8

Now, we can solve for our zeroes.

-4608/L^2 + 8 = 0

-8 = -8

-4608/L^2 = -8

-4608 = -8L^2

-4608/-8 = -8L^2/-8

576 = L^2

sqrt(576) = L

24 = L

So, the length of our printed area is 24 cm. Solving for the width of our printed area gives us 384/24 which is 16.

Okay, you are not done yet. You have only found the dimensions of the printed area, now you have to go back and add in the margins. The length had margins of 6 on top and bottom so we add 12 to 24 and get 36 cm. The width had margins on the left and right of 4 cm. each so we add 8 to the width of 16 and get 24 cm.

So, the overall dimensions of our poster are 36 cm. by 24 cm.

Let x and y the length and width of the printed portion on the poster. Using the formula for rectangle, the area of the printed portion is:

`A = xy`

`384=xy` ==> Let this be EQ.1 .

Then, take note of the margins of the poster.So, the dimensions of the poster are:

length: `x+8 ` width: `y+12`

The area of the poster is:

`A=(x+8)(y+12)`

Express the area as one variable. From EQ.1, replace y with 384/x.

`A=(x+8)(384/x+12) = 384 + 12x+3072/x+96`

`A=12x+3072/x+480`

To determine the dimensions of the poster that would result to minimum area, take the derivative of A.

`A' = 12-3072/x^2`

Set A'=0 and solve for x.

`0 = 12 - 3072/x^2`

`3072/x^2 = 12`

`x^2=3072/12`

`x^2=256`

`x=+-sqrt256 = +-16`

Since x is a dimension, take the positive value of x. Then, substitute it to EQ.1.

`384 =16y`

`y = 384/16 = 24`

Then, compute the length and width of the poster.

length: `x+8 = 16+8 = 24`

width: `y+12=24+12=36`

**Hence, a poster with smallest area has a dimension of 24cm x 36cm. **