Question
A rectangular picture frame is 5
inches wide and 10
inches tall. You want to make the area 6
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.
inches wide and 10
inches tall. You want to make the area 6
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.
Answers
GPT-4o mini
Let's denote the current width of the picture frame as \(w = 5\) inches and the current height as \(h = 10\) inches. The area \(A\) of the rectangular picture frame is given by:
\[
A = w \times h = 5 \times 10 = 50 \text{ square inches}
\]
We want to increase the area to 6 times the original area:
\[
\text{New Area} = 6A = 6 \times 50 = 300 \text{ square inches}
\]
Let \(x\) be the amount in inches by which we increase both the width and height. The new width will be \(w + x\) and the new height will be \(h + x\). Therefore, the equation for the new area becomes:
\[
(w + x)(h + x) = 300
\]
Substituting the known values of \(w\) and \(h\):
\[
(5 + x)(10 + x) = 300
\]
Now, we can expand the left-hand side:
\[
50 + 5x + 10x + x^2 = 300
\]
This simplifies to:
\[
x^2 + 15x + 50 = 300
\]
Next, we bring everything to one side of the equation:
\[
x^2 + 15x + 50 - 300 = 0
\]
This simplifies to:
\[
x^2 + 15x - 250 = 0
\]
Now we can apply the quadratic formula \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\) to solve for \(x\), where \(a = 1\), \(b = 15\), and \(c = -250\).
First, we calculate the discriminant:
\[
b^2 - 4ac = 15^2 - 4(1)(-250) = 225 + 1000 = 1225
\]
Now we can find \(x\):
\[
x = \frac{{-15 \pm \sqrt{1225}}}{2(1)} = \frac{{-15 \pm 35}}{2}
\]
This gives us two potential solutions:
1. \(x = \frac{{20}}{2} = 10\)
2. \(x = \frac{{-50}}{2} = -25\) (not a valid solution since \(x\) cannot be negative)
Thus, the only feasible solution is:
\[
x = 10
\]
Therefore, each dimension must be increased by **10 inches**.
\[
A = w \times h = 5 \times 10 = 50 \text{ square inches}
\]
We want to increase the area to 6 times the original area:
\[
\text{New Area} = 6A = 6 \times 50 = 300 \text{ square inches}
\]
Let \(x\) be the amount in inches by which we increase both the width and height. The new width will be \(w + x\) and the new height will be \(h + x\). Therefore, the equation for the new area becomes:
\[
(w + x)(h + x) = 300
\]
Substituting the known values of \(w\) and \(h\):
\[
(5 + x)(10 + x) = 300
\]
Now, we can expand the left-hand side:
\[
50 + 5x + 10x + x^2 = 300
\]
This simplifies to:
\[
x^2 + 15x + 50 = 300
\]
Next, we bring everything to one side of the equation:
\[
x^2 + 15x + 50 - 300 = 0
\]
This simplifies to:
\[
x^2 + 15x - 250 = 0
\]
Now we can apply the quadratic formula \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\) to solve for \(x\), where \(a = 1\), \(b = 15\), and \(c = -250\).
First, we calculate the discriminant:
\[
b^2 - 4ac = 15^2 - 4(1)(-250) = 225 + 1000 = 1225
\]
Now we can find \(x\):
\[
x = \frac{{-15 \pm \sqrt{1225}}}{2(1)} = \frac{{-15 \pm 35}}{2}
\]
This gives us two potential solutions:
1. \(x = \frac{{20}}{2} = 10\)
2. \(x = \frac{{-50}}{2} = -25\) (not a valid solution since \(x\) cannot be negative)
Thus, the only feasible solution is:
\[
x = 10
\]
Therefore, each dimension must be increased by **10 inches**.