We are given the perimeter \( P \) and the area \( A \) of a baby's crib, where:
- Perimeter \( P = 146 \) inches
- Area \( A = 1,276 \) square inches
Let the length of the crib be \( L \) inches and the width be \( W \) inches. The relationships for perimeter and area are given by:
\[ P = 2L + 2W \] \[ A = L \cdot W \]
Substituting the values we have:
-
From the perimeter: \[ 2L + 2W = 146 \] Dividing the whole equation by 2: \[ L + W = 73 \]
-
From the area: \[ L \cdot W = 1276 \]
Now, we can express \( W \) in terms of \( L \): \[ W = 73 - L \]
Next, substitute \( W \) into the area equation: \[ L(73 - L) = 1276 \] Expanding this gives us: \[ 73L - L^2 = 1276 \] Rearranging the equation leads to: \[ L^2 - 73L + 1276 = 0 \]
Now we can use the quadratic formula to solve for \( L \): \[ L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -73 \), and \( c = 1276 \). First, compute the discriminant: \[ b^2 - 4ac = (-73)^2 - 4 \cdot 1 \cdot 1276 \] \[ = 5329 - 5104 = 225 \]
Now, calculate \( L \): \[ L = \frac{73 \pm \sqrt{225}}{2} = \frac{73 \pm 15}{2} \]
Calculating the two potential values for \( L \):
- \( L = \frac{73 + 15}{2} = \frac{88}{2} = 44 \)
- \( L = \frac{73 - 15}{2} = \frac{58}{2} = 29 \)
Using \( L = 44 \) to find \( W \): \[ W = 73 - 44 = 29 \]
Using \( L = 29 \) to find \( W \): \[ W = 73 - 29 = 44 \]
Thus, the dimensions of the crib are: \[ \boxed{44 \text{ inches by } 29 \text{ inches}} \]