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### Special Cases and Applications

Question ID: TM6653

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Katie wants to create a rectangular frame for a picture. She has 60 inches of material. If she wants the length to be 3 inches more than 2 times the width, what is the largest possible length?

Write an equation and solve.

1. [tex]6w + 6 = 60[/tex]
2. [tex](2w + 3) \cdot 4 = 60[/tex]
3. [tex]4(2w + 3) = 60[/tex]
4. [tex]6w + 6 = 60[/tex]

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- Submit
- Pass
- Solve and close
- Don't know the answer

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Solution:

Let's write the equation based on the problem statement:

Given:
- Total material (perimeter) = 60 inches
- Length (L) = 2 times the width (W) plus 3 inches
- The perimeter formula for a rectangle is: [tex]P = 2L + 2W[/tex]

Step-by-step solution:

1. Write the equation for the perimeter:
[tex]\[
2L + 2W = 60
\][/tex]
2. Substitute [tex]L[/tex] with [tex]2W + 3[/tex]:
[tex]\[
2(2W + 3) + 2W = 60
\][/tex]
3. Simplify and solve for [tex]W[/tex]:
[tex]\[
4W + 6 + 2W = 60
\][/tex]
[tex]\[
6W + 6 = 60
\][/tex]
[tex]\[
6W = 54
\][/tex]
[tex]\[
W = 9
\][/tex]
4. Calculate [tex]L[/tex]:
[tex]\[
L = 2W + 3
\][/tex]
[tex]\[
L = 2(9) + 3
\][/tex]
[tex]\[
L = 18 + 3
\][/tex]
[tex]\[
L = 21
\][/tex]

So, the largest possible length is [tex]21[/tex] inches.


Sagot :

Sure, let's work through this problem together step-by-step.

Katic wants to create a rectangular frame for a picture using 60 inches of material. The length of the frame should be 3 inches more than 2 times its width.

Given:
1. Total material available for the perimeter of the frame: 60 inches.
2. The length ([tex]\( l \)[/tex]) is [tex]\( 3 \)[/tex] more than [tex]\( 2 \)[/tex] times the width ([tex]\( w \)[/tex]).

To find the largest possible length, follow these steps:

### Step-by-Step Solution:

1. Establish Variables:
- Let the width be [tex]\( w \)[/tex].
- According to the problem, the length [tex]\( l \)[/tex] is given by:
[tex]\[ l = 2w + 3 \][/tex]

2. Formulate the Perimeter Equation:
- The perimeter ([tex]\( P \)[/tex]) of a rectangle is given by the formula:
[tex]\[ P = 2(l + w) \][/tex]
- Substitute the given total material into the perimeter equation:
[tex]\[ 2(l + w) = 60 \][/tex]
- Substitute [tex]\( l = 2w + 3 \)[/tex] into the perimeter equation:
[tex]\[ 2((2w + 3) + w) = 60 \][/tex]

3. Simplify the Equation:
- Combine like terms within the parentheses:
[tex]\[ 2(2w + 3 + w) = 60 \][/tex]
- This simplifies to:
[tex]\[ 2(3w + 3) = 60 \][/tex]
- Distribute the 2:
[tex]\[ 6w + 6 = 60 \][/tex]

4. Solve for [tex]\( w \)[/tex] (width):
- Subtract 6 from both sides:
[tex]\[ 6w + 6 - 6 = 60 - 6 \][/tex]
[tex]\[ 6w = 54 \][/tex]
- Divide both sides by 6:
[tex]\[ w = \frac{54}{6} \][/tex]
[tex]\[ w = 9 \][/tex]

5. Find the Length ([tex]\( l \)[/tex]):
- Substitute [tex]\( w = 9 \)[/tex] back into the length equation:
[tex]\[ l = 2w + 3 \][/tex]
[tex]\[ l = 2(9) + 3 \][/tex]
[tex]\[ l = 18 + 3 \][/tex]
[tex]\[ l = 21 \][/tex]

### Conclusion:
The largest possible length of the rectangular frame that Katic can create, given 60 inches of material and the condition for the length being 3 more than 2 times the width, is [tex]\( \boxed{21} \)[/tex] inches.