Question
A rectangular picture frame is 6
inches wide and 10
inches tall. You want to make the area 7
times as large by increasing the length and width by the same amount.
Find the number of inches by which each dimension must be increased. Round to the nearest tenth.
inches wide and 10
inches tall. You want to make the area 7
times as large by increasing the length and width by the same amount.
Find the number of inches by which each dimension must be increased. Round to the nearest tenth.
Answers
GPT-4o mini
To find the increase in dimensions needed to make the area of the rectangular picture frame 7 times larger, we can follow these steps:
1. **Calculate the current area of the picture frame**:
The current dimensions of the picture frame are 6 inches wide and 10 inches tall.
\[
\text{Current Area} = \text{Width} \times \text{Height} = 6 \times 10 = 60 \text{ square inches}
\]
2. **Calculate the desired area**:
We want the new area to be 7 times the current area.
\[
\text{Desired Area} = 7 \times 60 = 420 \text{ square inches}
\]
3. **Set up the equation for the new dimensions**:
Let \( x \) be the increase in inches for both dimensions. The new width will be \( 6 + x \) and the new height will be \( 10 + x \).
The equation for the new area becomes:
\[
(6 + x)(10 + x) = 420
\]
4. **Expand and rearrange the equation**:
\[
60 + 6x + 10x + x^2 = 420
\]
\[
x^2 + 16x + 60 = 420
\]
\[
x^2 + 16x - 360 = 0
\]
5. **Solve the quadratic equation using the quadratic formula**:
The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 16 \), and \( c = -360 \).
First, calculate the discriminant:
\[
b^2 - 4ac = 16^2 - 4 \cdot 1 \cdot (-360) = 256 + 1440 = 1696
\]
Now, apply the quadratic formula:
\[
x = \frac{-16 \pm \sqrt{1696}}{2}
\]
Calculate \( \sqrt{1696} \):
\[
\sqrt{1696} \approx 41.2
\]
Therefore,
\[
x = \frac{-16 \pm 41.2}{2}
\]
This gives us two possible solutions:
\[
x = \frac{-16 + 41.2}{2} \approx \frac{25.2}{2} \approx 12.6
\]
\[
x = \frac{-16 - 41.2}{2} \approx \frac{-57.2}{2} \approx -28.6 \, (\text{not a valid solution since } x \text{ must be non-negative})
\]
6. **Final Answer**:
Thus, the increase in each dimension must be approximately **12.6 inches** (rounded to the nearest tenth).
1. **Calculate the current area of the picture frame**:
The current dimensions of the picture frame are 6 inches wide and 10 inches tall.
\[
\text{Current Area} = \text{Width} \times \text{Height} = 6 \times 10 = 60 \text{ square inches}
\]
2. **Calculate the desired area**:
We want the new area to be 7 times the current area.
\[
\text{Desired Area} = 7 \times 60 = 420 \text{ square inches}
\]
3. **Set up the equation for the new dimensions**:
Let \( x \) be the increase in inches for both dimensions. The new width will be \( 6 + x \) and the new height will be \( 10 + x \).
The equation for the new area becomes:
\[
(6 + x)(10 + x) = 420
\]
4. **Expand and rearrange the equation**:
\[
60 + 6x + 10x + x^2 = 420
\]
\[
x^2 + 16x + 60 = 420
\]
\[
x^2 + 16x - 360 = 0
\]
5. **Solve the quadratic equation using the quadratic formula**:
The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 16 \), and \( c = -360 \).
First, calculate the discriminant:
\[
b^2 - 4ac = 16^2 - 4 \cdot 1 \cdot (-360) = 256 + 1440 = 1696
\]
Now, apply the quadratic formula:
\[
x = \frac{-16 \pm \sqrt{1696}}{2}
\]
Calculate \( \sqrt{1696} \):
\[
\sqrt{1696} \approx 41.2
\]
Therefore,
\[
x = \frac{-16 \pm 41.2}{2}
\]
This gives us two possible solutions:
\[
x = \frac{-16 + 41.2}{2} \approx \frac{25.2}{2} \approx 12.6
\]
\[
x = \frac{-16 - 41.2}{2} \approx \frac{-57.2}{2} \approx -28.6 \, (\text{not a valid solution since } x \text{ must be non-negative})
\]
6. **Final Answer**:
Thus, the increase in each dimension must be approximately **12.6 inches** (rounded to the nearest tenth).