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Sagot :
To calculate the marginal cost of each quantity of bikes, we need to determine the additional cost incurred to produce one more bike compared to the previous quantity. This is done by subtracting the total cost of producing the previous quantity from the total cost of producing the current quantity.
Let's calculate the marginal costs step-by-step using the given table:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Bikes produced per day} & \text{Total cost} & \text{Marginal cost} \\ \hline 1 & \$80 & - \\ \hline 2 & \$97 & \$97 - \$80 = \$17 \\ \hline 3 & \$110 & \$110 - \$97 = \$13 \\ \hline 4 & \$130 & \$130 - \$110 = \$20 \\ \hline 5 & \$160 & \$160 - \$130 = \$30 \\ \hline 6 & \$210 & \$210 - \$160 = \$50 \\ \hline 7 & \$270 & \$270 - \$210 = \$60 \\ \hline \end{array} \][/tex]
Now let’s extract specific marginal costs of interest:
- The marginal cost of the first bike (additional cost to produce the 1st bike) is not defined using the typical method since it would be the cost of producing 0 bikes, which is not practical here.
- The marginal cost of the fourth bike can be found as the cost difference from the third to the fourth bike:
[tex]\[ \$130 - \$110 = \$20 \][/tex]
- The marginal cost of the sixth bike can be found as the cost difference from the fifth to the sixth bike:
[tex]\[ \$210 - \$160 = \$50 \][/tex]
- The marginal cost of the seventh bike can be found as the cost difference from the sixth to the seventh bike:
[tex]\[ \$270 - \$210 = \$60 \][/tex]
Thus, we can fill in the specifics:
- The marginal cost of the first bike: [tex]\( \$\text{N/A} \)[/tex]
- The marginal cost of the fourth bike: [tex]\( \$20 \)[/tex]
- The marginal cost of the sixth bike: [tex]\( \$50 \)[/tex]
- The marginal cost of the seventh bike: [tex]\( \$60 \)[/tex]
Let's calculate the marginal costs step-by-step using the given table:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Bikes produced per day} & \text{Total cost} & \text{Marginal cost} \\ \hline 1 & \$80 & - \\ \hline 2 & \$97 & \$97 - \$80 = \$17 \\ \hline 3 & \$110 & \$110 - \$97 = \$13 \\ \hline 4 & \$130 & \$130 - \$110 = \$20 \\ \hline 5 & \$160 & \$160 - \$130 = \$30 \\ \hline 6 & \$210 & \$210 - \$160 = \$50 \\ \hline 7 & \$270 & \$270 - \$210 = \$60 \\ \hline \end{array} \][/tex]
Now let’s extract specific marginal costs of interest:
- The marginal cost of the first bike (additional cost to produce the 1st bike) is not defined using the typical method since it would be the cost of producing 0 bikes, which is not practical here.
- The marginal cost of the fourth bike can be found as the cost difference from the third to the fourth bike:
[tex]\[ \$130 - \$110 = \$20 \][/tex]
- The marginal cost of the sixth bike can be found as the cost difference from the fifth to the sixth bike:
[tex]\[ \$210 - \$160 = \$50 \][/tex]
- The marginal cost of the seventh bike can be found as the cost difference from the sixth to the seventh bike:
[tex]\[ \$270 - \$210 = \$60 \][/tex]
Thus, we can fill in the specifics:
- The marginal cost of the first bike: [tex]\( \$\text{N/A} \)[/tex]
- The marginal cost of the fourth bike: [tex]\( \$20 \)[/tex]
- The marginal cost of the sixth bike: [tex]\( \$50 \)[/tex]
- The marginal cost of the seventh bike: [tex]\( \$60 \)[/tex]
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