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A café owner is designing a new menu and wants to include a decorative border around the outside of her food listings. Due to the cost of printing, the border should have an area of 48 square inches. The width of the border needs to be uniform around the entire menu. She has already determined that her food listings will fit within a 13-inch by 9-inch rectangular area.



The area of the decorative border can be modeled by the following equation, where x represents the width of the decorative border.

___x2 + ____ x = ____

Is it reasonable for the border to be 2.5 inches wide? ______(yes/no)

User Harumi
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2 Answers

4 votes

Answer:


\displaystyle 4x^2 + 44 = 48

Explanation:


\displaystyle [2x + 9][2x + 13] + 117 = 165 \\ \\ 4x^2 + 44x + 117 = 165 \\ \\ 4x^2 + 44x - 48 \\ [4x^2 + 48x] - [4x - 48] \\ 4x[x + 12] \: - 4[x + 12] \\ \\ [4x - 4][x + 12] \\ \\ -12, 1 = x

Since we are talking about LENGTHS, we only want the NON-NEGATIVE root, which in this case is 1, therefore the border is one inch wide, so 2½ inches is NOT reasonable.

I am joyous to assist you anytime.

User Jacenta
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2 votes

Answer:

  • 4x^2 +44x = 48
  • No, it is not reasonable

Explanation:

If x is the width of the border, it adds 2x to each food listing dimension. The overall area of the menu and border will be ...

total area = (13+2x)(9+2x) = 4x^2 +44x +117

The area of the food listing is ...

food listing area = (13)(9) = 117 . . . . square inches

So, the border area is the difference of these:

border area = total area - food listing area

border area = (4x^2 +44x +117) - (117)

border area = 4x^2 +44x

We want this to be 48 square inches, so a suitable model is ...

4x^2 +44x = 48

__

It is not reasonable for the border to be 2.5 inches wide.

_____

Why 2.5 inches is not reasonable

Dividing by 4 the equation becomes ...

x(x +11) = 12

One of the solutions is x=1. Using the value 2.5 for x would give ...

2.5(13.5) = 12 . . . . . false

User Brian Weiner
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5.8k points