Question

A rectangular hotel room has an area of 255 square feet and a perimeter of 64 feet. What are the dimensions of the room?

feet by
feet

Answer 1

Let the length of the room be \( l \) and the width be \( w \).

We have the following two equations based on the area and perimeter:

1. Area equation:
\[
l \cdot w = 255
\]

2. Perimeter equation:
\[
2l + 2w = 64
\]

We can simplify the perimeter equation:
\[
l + w = 32
\]

From this equation, we can express \( w \) in terms of \( l \):
\[
w = 32 - l
\]

Now we can substitute \( w \) into the area equation:
\[
l(32 - l) = 255
\]
\[
32l - l^2 = 255
\]
Rearranging the equation gives us:
\[
l^2 - 32l + 255 = 0
\]

Now, we can use the quadratic formula to solve for \( l \):
\[
l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = -32 \), and \( c = 255 \).

Calculating the discriminant:
\[
b^2 - 4ac = (-32)^2 - 4 \cdot 1 \cdot 255 = 1024 - 1020 = 4
\]

Now substituting back into the quadratic formula:
\[
l = \frac{32 \pm \sqrt{4}}{2} = \frac{32 \pm 2}{2}
\]

This gives us two possible solutions for \( l \):
\[
l = \frac{34}{2} = 17 \quad \text{or} \quad l = \frac{30}{2} = 15
\]

So, the dimensions are:
1. \( l = 17 \) and \( w = 32 - 17 = 15 \)
2. \( l = 15 \) and \( w = 32 - 15 = 17 \)

Thus, the dimensions of the room are:
\[
\boxed{17 \text{ feet by } 15 \text{ feet}}
\]