Question

19. Given the total cost function
TC = 150 Q – 3Q2 + 0.25Q3
Complete the following table by computing the total, average, and marginal costs associated with each quantity indicated.

Quantity Total Cost Average Cost Marginal Cost
1
2
3
4
5
6

Answer 1

We are given:

$TC = 150Q - 3Q^2+0.25Q^3$

The average cost, $AC = \dfrac {TC}{Q}$ and,

Marginal cost, $MC$ is such that

$MC_{n} = TC_{n}-TC_{n-1}$

$\bold {Example}$

When $Q = 1$,

$TC = 150(1)-3(1^2)+0.25(1^3)$

$\bold {= \$147.25}$

$AC = \dfrac {\$147.25}{1}$ $\bold {=\$147.25}$

$MC = TC_{1}-TC_{0}$

$\bold {=?}$ ($TC_{0}$ does not exist)

When $Q=2,$

$TC = 150(2)-3(2^2)+0.25(2^3)$

$\bold {=\$290}$

$AC = \dfrac {\$290}{2}$ $\bold {=\$145}$

$MC = TC_{2}-TC_{1}$

$= \$290-\$147.25$

$\bold {=\$142.75}$

When $Q=3$

$TC = 150(3)-3(3^2)+0.25(3^3)$

$\bold {=\$429.75}$

$AC = \dfrac {\$429.75}{3}$ $\bold {=\$143.25}$

$MC = TC_{3}-TC_{2}$

$= \$429.75-\$290$

$\bold {=\$139.75}$

The remaining Q=4, 5 and 6 are completed in a similar manner. The table below is the completed solution.

$\bold {ANSWER}$

$\\ \bold {Q \hspace{9mm} TC \hspace {11mm} AC \hspace {11mm} MC}$

$1 \hspace{8mm} 147.25 \hspace {8mm} 147.25 \hspace {10mm} -$

$2 \hspace{8mm} 290.00 \hspace {8mm} 145.00\hspace {8mm} 142.75$

$3 \hspace{8mm} 429.75 \hspace {8mm} 143.25 \hspace {8mm} 139.75$

$4\hspace{8mm} 568.00 \hspace {8mm} 142.00 \hspace {8mm} 138.25$

$5 \hspace{8mm} 706.25 \hspace {8mm} 141.25 \hspace {8mm} 138.25$

$6 \hspace{8mm} 846.00 \hspace {8mm} 141.00 \hspace {8mm} 139.75$