←
Modeling a Quadratic Problem

8.1.2 Modeling a Quadratic Problem

8.1.2 • Modeling a Quadratic Problem

Activity

Your teacher will give you a picture that is 7 inches by 4 inches, a piece of colored paper that measures 4 inches by 2.5 inches representing framing material, a ruler, and a pair of scissors.

Cut the framing material to create a rectangular frame for the picture. The frame should have the same width all the way around and have no overlaps. All of the framing material should be used with no leftover pieces. The frame should not cover any part of the picture.

Let’s review the information we know.

  • The picture measures 7 inches by 4 inches.

  • There are 10 square inches of framing material.

  • The frame must be uniform in width.

  • All of the framing material must be used.

  • The frame should not cover any part of the picture.

Describe the process you used to create the frame.

Compare your answer:

Here are some examples of the process shown in images. Only some of the attempts are correct while others are not.

A collage of five images showing the steps of gluing and folding strips of paper around the edges of rectangular pieces of paper, demonstrating a paper-framing or matting technique on wooden and white backgrounds.

Now that you have completed the activity, can you think of any mathematical method to solve the problem?

Are you ready for more?

Extending Your Thinking

Juan says, “The perimeter of the picture is 22 inches. If I cut the framing material into 9 pieces, each one being 2.5 inches by 4949 inch, I’ll have more than enough material to surround the picture because those pieces would yield 9·2.59·2.5, or 22.5 inches for the frame.”

Do you agree with Juan? Be prepared to show your reasoning.

Compare your answer:

No, Juan is not correct because 9 pieces would not be enough to go around the frame. While the total length of those pieces is enough to surround the picture, he will need at least two pieces for each short side to cover 4 inches and at least 3 pieces for each long side to cover 7 inches. This adds up to 10 pieces, as shown below.

A rectangular landscape painting, 4 inches by 7 inches, with a sun and mountains, will be framed with strips each 2.5 inches long and 4/9 inch wide, forming a border around the painting.

Self Check

Suppose you want to frame a picture that measures 8 inches by 3 inches. The frame is created from 12 square inches of framing material. The frame must be uniform in width, and all of the framing material must be used. Which statement is true?
  1. The total area of the framed picture will be 36 square inches.
  2. The perimeter of the framed picture will be 22 inches.
  3. The total area of the framed picture will be 24 square inches.
  4. The frame will measure exactly 8 inches by 3 inches.

Additional Resources

Modeling a Quadratic Problem

This activity challenges you to complete a very difficult task!

Let’s review the information we know from the picture in the Self Check.

  • The picture measures 8 inches by 3 inches.
  • There are 12 square inches of framing material.
  • The frame must be uniform in width.
  • All of the framing material must be used.
  • The frame should not cover any part of the picture.

What other information can you learn from the situation?

  • The perimeter of the picture is 8+8+3+38+8+3+3, or 22 inches. The framed picture will have a perimeter greater than this.
  • If the picture is 8 inches by 3 inches, then the area of the picture is 8·38·3, or 24 square inches.
  • Since there are 12 square inches of framing material and all of it must be used, the framed picture will have an area of 24+1224+12, or 36 square inches.

The more you know about the situation, the closer you are to being able to solve this problem mathematically instead of guessing and checking to see if you are correct.

We will use this information in later activities to set up a quadratic equation that can be used to solve the problem.

Try it

Modeling a Quadratic Problem

Suppose you want to frame a different picture that measures 10 inches by 6 inches. The frame is created from 36 square inches of framing material. The frame must be uniform in width, and all of the framing material must be used.

1. What is the perimeter of the picture?

2. What is the area of the picture?

3. What will be the total area of the framed picture?

What is the perimeter of the picture?

Here is how to find the missing information about the framed picture:

The perimeter of the picture is 10+10+6+6=3210+10+6+6=32 inches.

Suppose you want to frame a different picture that measures 10 inches by 6 inches. The frame is created from 36 square inches of framing material. The frame must be uniform in width, and all of the framing material must be used.

4. What will be the total area of the framed picture?

Here is how to find the missing information about the framed picture:

The total area of the framed picture will be 60+36=9660+36=96 square inches.

5. What is the area of the picture?

Here is how to find the missing information about the framed picture:

The area of the picture is 10·6=6010·6=60 square inches.