Katie wants to create a rectangular frame for a picture using 60 inches of material. She also wants the length of the frame to be 3 inches more than 2 times the width. Let's determine the dimensions of the frame step-by-step.
Let's denote the width of the frame as [tex]\( w \)[/tex] (in inches).
According to the problem, the length of the frame, [tex]\( l \)[/tex], is:
[tex]\[ l = 2w + 3 \][/tex]
The total length of the material used for the perimeter of the rectangle is 60 inches. The perimeter [tex]\( P \)[/tex] of a rectangle can be found using the formula:
[tex]\[ P = 2 \cdot (\text{length} + \text{width}) \][/tex]
Using Katie’s conditions, this becomes:
[tex]\[ 60 = 2 \cdot (l + w) \][/tex]
Substituting the expression for [tex]\( l \)[/tex] into this equation, we get:
[tex]\[ 60 = 2 \cdot ((2w + 3) + w) \][/tex]
Simplify inside the parentheses:
[tex]\[ 60 = 2 \cdot (3w + 3) \][/tex]
Distribute the 2:
[tex]\[ 60 = 6w + 6 \][/tex]
Next, to find [tex]\( w \)[/tex], isolate it by following these steps:
1. Subtract 6 from both sides:
[tex]\[ 60 - 6 = 6w \][/tex]
[tex]\[ 54 = 6w \][/tex]
2. Divide by 6:
[tex]\[ w = \frac{54}{6} \][/tex]
[tex]\[ w = 9 \][/tex]
Thus, the width [tex]\( w \)[/tex] is 9 inches.
Now, to find the length [tex]\( l \)[/tex], substitute [tex]\( w = 9 \)[/tex] back into the equation for [tex]\( l \)[/tex]:
[tex]\[ l = 2w + 3 \][/tex]
[tex]\[ l = 2(9) + 3 \][/tex]
[tex]\[ l = 18 + 3 \][/tex]
[tex]\[ l = 21 \][/tex]
Hence, the width is 9 inches and the largest possible length is 21 inches.
