# Page Design

• November 4th 2008, 07:30 AM
magentarita
Page Design
A page with dimensions of (17/2) inches by 11 inches has a border of uniform width x surrounding the printed matter of the page.

(a) Write a formula for the area A of the printed part of the page as a function of the width x of the border.

(b) Give the domain and range of the area A.

(c) Find the area of the printed page for borders of widths 1 inch, 1.2 inches and 1.5 inches.

• November 5th 2008, 02:01 PM
masters
Quote:

Originally Posted by magentarita
A page with dimensions of (17/2) inches by 11 inches has a border of uniform width x surrounding the printed matter of the page.

(a) Write a formula for the area A of the printed part of the page as a function of the width x of the border.

(b) Give the domain and range of the area A.

(c) Find the area of the printed page for borders of widths 1 inch, 1.2 inches and 1.5 inches.

Area of paper = $11\left(\frac{17}{2}\right)=\frac{187}{2} \ \ in^2$

Area of the printed part = $A(x)=\left(\frac{17}{2}-2x\right)\left(11-2x\right)=\frac{187}{2}-39x+4x^2$

Area of the printed part cannot exceed area of the paper, so

$4x^2-39x+\frac{187}{2}<\frac{187}{2}$

Domain of Area A = $\left\{0

Range of Area A = $\left\{0

For part (c), just substitute these values in A(x).
• November 5th 2008, 03:17 PM
magentarita
ok...........
Quote:

Originally Posted by masters
Area of paper = $11\left(\frac{17}{2}\right)=\frac{187}{2} \ \ in^2$

Area of the printed part = $A(x)=\left(\frac{17}{2}-2x\right)\left(11-2x\right)=\frac{187}{2}-39x+4x^2$

Area of the printed part cannot exceed area of the paper, so

$4x^2-39x+\frac{187}{2}<\frac{187}{2}$

Domain of Area A = $\left\{0

Range of Area A = $\left\{0

For part (c), just substitute these values in A(x).

As always, I thank you for your great help.