Answer to Question #172438 in Microeconomics for yaashi mehta

Question

Output AC. AFC. AVC. MC

1. 100. 40

2. 36

3. 36

4. 36

5. 12. 41

Answer 1

Let's first find the fixed cost (FC):

$AFC_1=\dfrac{FC}{Q_1},$ $FC=AFC_1Q_1=\$40\cdot1=\$40.$

Then, we can find $AFC$ for the output from $Q=2$ to $Q=5$:

$AFC_2=\dfrac{FC}{Q_2}=\dfrac{\$40}{2}=\$20,$ $AFC_3=\dfrac{FC}{Q_3}=\dfrac{\$40}{3}=\$13.33,$ $AFC_4=\dfrac{FC}{Q_4}=\dfrac{\$40}{4}=\$10,$ $AFC_5=\dfrac{FC}{Q_5}=\dfrac{\$40}{5}=\$8.$

Since, $AC=AFC+AVC$, we can find $AVC$ for each number of output:

$AVC_1=AC_1-AFC_1=\$100-\$40=\$60,$ $AVC_2=AC_2-AFC_2=\$36-\$20=\$16,$ $AVC_3=AC_3-AFC_3=\$36-\$13.33=\$22.67,$ $AVC_4=AC_4-AFC_4=\$36-\$10=\$26,$ $AVC_5=AC_5-AFC_5=\$12-\$8=\$4.$

Let's find the TC. By the definition,

$AC=\dfrac{TC}{Q},$ $TC_1=AC_1\cdot Q_1=\$100\cdot1=\$100,$ $TC_2=AC_2\cdot Q_2=\$36\cdot2=\$72,$ $TC_3=AC_3\cdot Q_3=\$36\cdot3=\$108,$ $TC_4=AC_4\cdot Q_4=\$36\cdot4=\$144,$ $TC_5=AC_5\cdot Q_5=\$12\cdot5=\$60.$

By the definition of MC we have:

$MC_1=\dfrac{TC_2-TC_1}{Q_2-Q_1}=\dfrac{\$72-\$100}{2-1}=-\$28,$ $MC_2=\dfrac{TC_3-TC_2}{Q_3-Q_2}=\dfrac{\$108-\$72}{3-2}=\$36,$ $MC_3=\dfrac{TC_4-TC_3}{Q_4-Q_3}=\dfrac{\$144-\$108}{4-3}=\$36,$ $MC_4=\dfrac{TC_5-TC_4}{Q_5-Q_4}=\dfrac{\$60-\$144}{5-4}=-\$84.$

Finally, we get the following table: