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:
-
Area equation:
\[ l \cdot w = 255 \] -
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:
- \( l = 17 \) and \( w = 32 - 17 = 15 \)
- \( l = 15 \) and \( w = 32 - 15 = 17 \)
Thus, the dimensions of the room are: \[ \boxed{17 \text{ feet by } 15 \text{ feet}} \]