The printed area of a rectangular poster is 704 sq inches... the printed area plus the area of the margins is 1200 sq inches find the dimension of the poster if each of the margins is 4" wide...

:D !

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- Oct 3rd 2006, 10:05 AM^_^Engineer_Adam^_^Word problems involving Quadratic systems...
The printed area of a rectangular poster is 704 sq inches... the printed area plus the area of the margins is 1200 sq inches find the dimension of the poster if each of the margins is 4" wide...

:D ! - Oct 3rd 2006, 02:31 PMQuick
**704=w*l**

Adding the printed margins we get:**(w+8)*(l+8)=1200**

Use Foil:**w*l+8(w+l)+64=1200**

subtract 64 from both sides:**w*l+8(w+l)=1136**

Subtract this equation by the first equation:**w*l+8(w+l)-w*l=1136-704**

thus:**8(w+l)=432**

divide:**w+l=54**

Now you can choose between literally infinite possibilities for dimensions.

Edit: Apparently not :( I hope you haven't turned this in Adam... - Oct 3rd 2006, 10:08 PMSoroban
Hello, ^_^Engineer_Adam^_^!

Quote:

The printed area of a rectangular poster is 704 in².

The printed area plus the area of the margins is 1200 in².

Find the dimension of the poster if each of the margins is 4" wide.

Code:`: 4 : - x - : 4 :`

- * - - - - - - - - * -

4 | | :

- | * - - - - * | :

: | | | | :

: | | | | :

y | |y | | y+8

: | | | | :

: | | x | | :

- | * - - - - * | :

4 | | :

- * - - - - - - - - * -

: - - - x+8 - - - :

The length of the printed area is*x.*

The height of the printed area is*y.*

. . Then: .xy = 704 . → . y = 704/x**[1]**

The length of the poster is*x + 8.*

The height of the poster is*y + 8.*

. . Then: .(x + 8)(y + 8) = 1200**[2]**

Substitute [1] into [2]: .(x + 8)(704/x + 8) .= .1200

. . which simplifies to: .8x² - 432x + 5632 .= .0

. . Divide by 8: . x² - 54x + 704 .= .0

. . which factors: .(x - 22)(x - 32) .= .0

. . and has roots: .x .= .22, 32 . → . y .= .32, 22

Hence, the dimensions of the printed area is: 22 x 32 inches.

Therefore, the dimension of the poster is:**30 x 40 inches**.