The perimeter of the front of a 7-ounce Jiffy Muffin Mix box is 44 centimeters. The height of the front is 3 centimeters less than twice the width. What are the height and width of the box front? (Hint: Draw a diagram.)



Answer :

To find the dimensions of the box, namely the height and width, we will set up and solve an algebraic equation step by step.

1. Define Variables:
- Let [tex]\( w \)[/tex] represent the width of the box in centimeters.
- Let [tex]\( h \)[/tex] represent the height of the box in centimeters.

2. Write the Given Relationships:
- The perimeter of the front of the box is 44 centimeters.
- The height [tex]\( h \)[/tex] is 3 centimeters less than twice the width [tex]\( w \)[/tex].

3. Express the Height in Terms of Width:
- According to the problem, the height [tex]\( h \)[/tex] is 3 centimeters less than twice the width:
[tex]\[ h = 2w - 3 \][/tex]

4. Write the Perimeter Formula:
- The perimeter [tex]\( P \)[/tex] of the front of the box can be expressed in terms of width and height as follows:
[tex]\[ 2(w + h) = P \][/tex]
- Substituting the given perimeter [tex]\( P = 44 \)[/tex] centimeters into the equation gives:
[tex]\[ 2(w + h) = 44 \][/tex]

5. Substitute the Expression for [tex]\( h \)[/tex] into the Perimeter Formula:
- Substitute [tex]\( h = 2w - 3 \)[/tex] into the equation [tex]\( 2(w + h) = 44 \)[/tex]:
[tex]\[ 2(w + (2w - 3)) = 44 \][/tex]

6. Simplify the Equation:
- Combine like terms inside the parentheses:
[tex]\[ 2(w + 2w - 3) = 44 \][/tex]
[tex]\[ 2(3w - 3) = 44 \][/tex]
- Distribute the 2:
[tex]\[ 6w - 6 = 44 \][/tex]

7. Solve for [tex]\( w \)[/tex]:
- Add 6 to both sides of the equation:
[tex]\[ 6w - 6 + 6 = 44 + 6 \][/tex]
[tex]\[ 6w = 50 \][/tex]
- Divide both sides by 6:
[tex]\[ w = \frac{50}{6} = \frac{25}{3} \][/tex]

8. Find the Height [tex]\( h \)[/tex]:
- Use the relationship [tex]\( h = 2w - 3 \)[/tex] and substitute [tex]\( w = \frac{25}{3} \)[/tex]:
[tex]\[ h = 2\left(\frac{25}{3}\right) - 3 \][/tex]
[tex]\[ h = \frac{50}{3} - 3 \][/tex]
[tex]\[ h = \frac{50}{3} - \frac{9}{3} \][/tex]
[tex]\[ h = \frac{41}{3} \][/tex]

Thus, the width of the box is [tex]\(\frac{25}{3}\)[/tex] centimeters, and the height of the box is [tex]\(\frac{41}{3}\)[/tex] centimeters.

Other Questions