Question

read and try to solve the problem below.
dalla bakes cornbread for family barbecues. the cornbread batter has a volume of 134 in.³. she needs at least 1\frac{1}{4} in. of space between the top of the batter and the top of the pan. will the batter fit in a rectangular pan that is 8 in. long, 12\frac{1}{2} in. wide, and 3 in. high?

Explanation:

Step1: Calculate the base area of the pan

The pan is rectangular, so the base area \( A \) is length \( l \) times width \( w \). Given \( l = 8 \) in and \( w = 12\frac{1}{2}=\frac{25}{2} \) in. So \( A = 8\times\frac{25}{2}= 100 \) in².

Step2: Find the maximum volume the pan can hold with the required space

The required space between the batter and the top is at least \( 1\frac{1}{4}=\frac{5}{4} \) in, and the pan's height available for batter (after considering the space) is \( 3 - \frac{5}{4}=\frac{12 - 5}{4}=\frac{7}{4} \) in? Wait, no, wait. Wait, the pan's height is 3 in? Wait, the problem says "a rectangular pan that is 8 in. long, 12 1/2 in. wide, and 3 in. high"? Wait, maybe I misread. Wait, the batter needs at least \( 1\frac{1}{4} \) in of space between the top of the batter and the top of the pan. So the height the batter can occupy is \( h_{batter}=3 - 1\frac{1}{4}=3-\frac{5}{4}=\frac{12 - 5}{4}=\frac{7}{4}=1.75 \) in? Wait, no, wait, maybe the pan's height is such that the available height for batter (before considering the space) is? Wait, no, the batter's volume is 134 in³. Wait, let's recast. The volume the pan can hold for the batter (with the required space) is base area times (pan height - required space). Wait, the pan's dimensions: length 8 in, width \( 12\frac{1}{2} \) in, height 3 in. The required space between batter and top is \( 1\frac{1}{4} \) in, so the maximum height the batter can have is \( 3 - 1\frac{1}{4}=1\frac{3}{4} \) in? Wait, no, \( 3 - 1\frac{1}{4}= \frac{12}{4}-\frac{5}{4}=\frac{7}{4}=1.75 \) in. Then the maximum volume the batter can occupy (to have the required space) is base area times this height. Base area is \( 8\times12\frac{1}{2}=8\times\frac{25}{2}=100 \) in². So maximum volume \( V_{max}=100\times1.75 = 175 \) in³.

Step3: Compare the batter's volume with \( V_{max} \)

The batter's volume is 134 in³. Since \( 134<175 \), the batter will fit. Wait, wait, maybe I messed up the height. Wait, maybe the pan's height is not 3 in? Wait, the problem says "a rectangular pan that is 8 in. long, 12 1/2 in. wide, and 3 in. high"? Wait, the question is "Will the batter fit in a rectangular pan that is 8 in. long, 12 1/2 in. wide, and 3 in. high?" with the batter needing at least \( 1\frac{1}{4} \) in of space between top of batter and top of pan. So the height the batter can take is \( 3 - 1\frac{1}{4}=1\frac{3}{4} \) in. Then the volume the batter would occupy in the pan (if it fits) is length × width × (height of batter). Wait, no, the batter's volume is 134 in³. Let's calculate the height the batter would have in the pan: \( h=\frac{V_{batter}}{A}=\frac{134}{100}=1.34 \) in. Then the space between batter and top is \( 3 - 1.34 = 1.66 \) in, which is more than \( 1\frac{1}{4}=1.25 \) in. So yes, it will fit. Wait, let's do it properly.

Wait, the pan's base area: \( l = 8 \) in, \( w = 12.5 \) in, so \( A = 8\times12.5 = 100 \) in². The batter's volume is \( V = 134 \) in³. The height the batter will take in the pan is \( h=\frac{V}{A}=\frac{134}{100}=1.34 \) in. The required minimum space between batter and top is \( 1.25 \) in. The total height used by batter and space is \( 1.34 + 1.25 = 2.59 \) in, which is less than the pan's height of 3 in. So yes, the batter will fit.

Wait, maybe the steps were:

1. Calculate the base area of the pan: \( 8 \times 12\frac{1}{2}=8\times\frac{25}{2}=100 \) in².

2. Calculate the height the batter would occupy: \( \frac{134}{100}=1.34 \) in.

3. Calculate the space between batter and top: \( 3 - 1.34 = 1.66 \) in.

4. Compare the space (1.66 in) with the required minimum (1.25 in). Since \( 1.66>1.25 \), the batter will fit.

Answer:

Yes, the batter will fit.