Answer to Question #142332 in Finance for Mr. Dinesh Pal Singh

Question #142332
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
1
Expert's answer
2020-11-10T07:24:07-0500

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"

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