Gerald owns a livestock trailer that can hold up to 10,000 pounds of livestock. He needs to transport 36 goats and 18 calves in his livestock trailer in one trip. Assume the trailer has enough room to transport this many livestock.

Question

Let g represent the average weight of each goat in pounds and let c represent the average weight of each calf in pounds.

Create the inequality that symbolizes this situation, and then answer the question below.

g +
c ≤

Can Gerald transport these goats and calves in his livestock trailer in one trip if the average weight of each goat is 96 pounds and the average weight of each calf is 369 pounds?

Answer 1

To create the inequality for this situation, we need to calculate the total weight of the goats and calves being transported. Given that there are 36 goats and 18 calves, the total weight can be represented as:

\[ 36g + 18c \leq 10,000 \]

Where:

• \(g\) is the average weight of each goat in pounds,
• \(c\) is the average weight of each calf in pounds,
• \(10,000\) is the maximum weight capacity of the trailer.

Now, we will plug in the average weights of the goats and calves:

• Average weight of each goat, \(g = 96\) pounds
• Average weight of each calf, \(c = 369\) pounds

Substituting the values into the inequality:

\[ 36(96) + 18(369) \leq 10,000 \]

Calculating the total weight:

1. Calculate the total weight of the goats: \[ 36 \times 96 = 3456 \text{ pounds} \]

2. Calculate the total weight of the calves: \[ 18 \times 369 = 6642 \text{ pounds} \]

Now, sum the weights of the goats and calves:

\[ 3456 + 6642 = 10098 \text{ pounds} \]

Now, we check the inequality:

\[ 10098 \leq 10,000 \]

Since \(10098\) is greater than \(10,000\), this means Gerald cannot transport these goats and calves in one trip because the total weight exceeds the trailer's capacity.

In summary, the inequality is:

\[ 36g + 18c \leq 10,000 \]

And based on the average weights given, Gerald cannot transport all the livestock in one trip.