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.
