Question

A baby's crib has a perimeter of 146 inches and an area of 1,276 square inches. What are the dimensions of the crib?

inches by
inches

Answer 1

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:

1. From the perimeter:
\[
2L + 2W = 146
\]
Dividing the whole equation by 2:
\[
L + W = 73
\]

2. 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 \( L \):
1. \( L = \frac{73 + 15}{2} = \frac{88}{2} = 44 \)
2. \( L = \frac{73 - 15}{2} = \frac{58}{2} = 29 \)

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:
\[
\boxed{44 \text{ inches by } 29 \text{ inches}}
\]