Question
A rectangular piece of art is 12 inches wide and 18 inches tall. A rectangular frame of uniform width surrounds the entire art piece. The total area of the art piece and frame is 352 square inches.
Which equation and solution represent w, the width of the frame in inches?
Which equation and solution represent w, the width of the frame in inches?
Answers
GPT-4o mini
To find the width \( w \) of the frame surrounding the artwork, we first need to determine the dimensions of the entire piece of art plus the frame.
The dimensions of the artwork are:
- Width = 12 inches
- Height = 18 inches
Since the frame surrounds the artwork with a uniform width \( w \), the overall dimensions of the artwork plus the frame will be:
- Overall width = \( 12 + 2w \) (because the frame adds \( w \) to both sides)
- Overall height = \( 18 + 2w \) (because the frame adds \( w \) to the top and \( w \) to the bottom)
The total area of the artwork plus the frame is given as 352 square inches. Therefore, we can set up the following equation for the area:
\[
(12 + 2w)(18 + 2w) = 352
\]
Now, we need to simplify and solve this equation.
1. Expand the left-hand side:
\[
(12 + 2w)(18 + 2w) = 12 \cdot 18 + 12 \cdot 2w + 18 \cdot 2w + 4w^2
\]
Calculating \( 12 \cdot 18 \):
\[
12 \cdot 18 = 216
\]
Thus, we have:
\[
216 + 24w + 36w + 4w^2 = 216 + 60w + 4w^2
\]
Therefore, our equation simplifies to:
\[
4w^2 + 60w + 216 = 352
\]
2. Subtract 352 from both sides:
\[
4w^2 + 60w + 216 - 352 = 0
\]
\[
4w^2 + 60w - 136 = 0
\]
3. This can be simplified by dividing everything by 4:
\[
w^2 + 15w - 34 = 0
\]
Now we have a quadratic equation:
\[
w^2 + 15w - 34 = 0
\]
To solve for \( w \), we can use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 15 \), and \( c = -34 \).
4. Calculate the discriminant:
\[
b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot (-34) = 225 + 136 = 361
\]
5. Now plug into the quadratic formula:
\[
w = \frac{-15 \pm \sqrt{361}}{2 \cdot 1}
\]
\[
w = \frac{-15 \pm 19}{2}
\]
6. This gives us two potential solutions:
- \( w = \frac{4}{2} = 2 \)
- \( w = \frac{-34}{2} = -17 \) (which we discard since width cannot be negative)
Thus, the solution for the width of the frame is:
\[
w = 2 \, \text{inches}
\]
The equation that leads to this solution is:
\[
w^2 + 15w - 34 = 0
\]
The dimensions of the artwork are:
- Width = 12 inches
- Height = 18 inches
Since the frame surrounds the artwork with a uniform width \( w \), the overall dimensions of the artwork plus the frame will be:
- Overall width = \( 12 + 2w \) (because the frame adds \( w \) to both sides)
- Overall height = \( 18 + 2w \) (because the frame adds \( w \) to the top and \( w \) to the bottom)
The total area of the artwork plus the frame is given as 352 square inches. Therefore, we can set up the following equation for the area:
\[
(12 + 2w)(18 + 2w) = 352
\]
Now, we need to simplify and solve this equation.
1. Expand the left-hand side:
\[
(12 + 2w)(18 + 2w) = 12 \cdot 18 + 12 \cdot 2w + 18 \cdot 2w + 4w^2
\]
Calculating \( 12 \cdot 18 \):
\[
12 \cdot 18 = 216
\]
Thus, we have:
\[
216 + 24w + 36w + 4w^2 = 216 + 60w + 4w^2
\]
Therefore, our equation simplifies to:
\[
4w^2 + 60w + 216 = 352
\]
2. Subtract 352 from both sides:
\[
4w^2 + 60w + 216 - 352 = 0
\]
\[
4w^2 + 60w - 136 = 0
\]
3. This can be simplified by dividing everything by 4:
\[
w^2 + 15w - 34 = 0
\]
Now we have a quadratic equation:
\[
w^2 + 15w - 34 = 0
\]
To solve for \( w \), we can use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 15 \), and \( c = -34 \).
4. Calculate the discriminant:
\[
b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot (-34) = 225 + 136 = 361
\]
5. Now plug into the quadratic formula:
\[
w = \frac{-15 \pm \sqrt{361}}{2 \cdot 1}
\]
\[
w = \frac{-15 \pm 19}{2}
\]
6. This gives us two potential solutions:
- \( w = \frac{4}{2} = 2 \)
- \( w = \frac{-34}{2} = -17 \) (which we discard since width cannot be negative)
Thus, the solution for the width of the frame is:
\[
w = 2 \, \text{inches}
\]
The equation that leads to this solution is:
\[
w^2 + 15w - 34 = 0
\]