A graphics designer is designing an advertising brochure for an art show

Question

A graphics designer is designing an advertising brochure for an art show. Each page of the brochure is rectangular with an area of 56in^2 and a perimeter of 30 in. Find the dimensions of the brochure.

The longer side is ___inches,

The short side is ____ inches.

Answer 1

Let W = width in inches

Let L = length in inches

PERIMETER = 30 inches

2W + 2L = 30

AREA = 56 SQUARE INCHES

LW = 56

L = 56/W

Substituting 56/W into the equation 2W + 2L = 30.

2W + (2)(56/W) = 30

2W + (112/W) = 30

2W^2 + 112 = 30W Multiply entire equation by "W".

2W^2 - 30W + 112 = 0

W^2 - 15W + 56 = 0 Divide entire equation by 2. Some say multiply by 1/2. Same thing.

(W - 8) (W - 7) = 0

So, W = 8 FT OR W = 7 FT

Since the WIDTH is always shorter than the LENGTH, the WIDTH IS 7 INCHES.

The longer measurement, 8 INCHES IS THE LENGTH.

CHECKING:

AREA 7 in X 8 in = 56 sq in

PERIMETER = 2W + 2L = (2)(7) + (2)(8) = 14 in + 16 in = 30 in

Hope this helps you TA.