alright so i sorta understand anx^n1 and getting the answer much more simple then i thought but can anyone tell me why vc is 2Q^2? and ATC = 8/Q+2Q
i dont understand derivatives i know how to calculate point price elasticity but thats only because i know its change in b x P/Q and not because i understand how derivatives work. please dumb it down for me.
i have a mid sem exam on Wednesday and would appreciate any help. also why is average total cost 8/q + 2Q. i understand 8 is the fixed cost and therefore 8/quantity makes average fixed cost. Likewise 2Q^2/Q makes average variable cost but why is the square sign gone. thanks everyone who can help i 100% appreciate any effort.
Consider the following algebraic example:
TC = 8 + 2Q^2
=

Calculate FC, VC, ATC and MC.

Answer:

FC = 8

VC = 2Q^2
2

ATC = 8/Q + 2Q

MC =
dTC/dQ
= 4Q (slope of TC curve)
Let T = total cost
Q = quantity of units produced
F = fixed costs, which that do not vary with quantity produced
V(Q) = variable costs, which do vary with quantity produced
A = average cost
M = marginal cost, being the incremental cost of one unit given that Q units are already being produced
OK with that notation?
We say that T = g(Q), meaning that the total cost varies in some well defined way as the number of units produced varies.
Then you are taught that M = g'(Q), where g'(Q) is the derivative of g(Q). Strictly speaking, this is a useful approximation only if Q is large, but economists don't teach that. In fact, they don't even admit it. So now I probably have the economists mad at me.
I am going to give you a "dumbed down" explanation of the derivative. This will probably get the mathematicians mad at me.
We shall use your problem as a concrete example.
$T = f(Q) = 8 + 2Q^2.$ That is a given. f(Q) is an abstract way of saying that cost varies with quantity produced. The problem gives a concrete example.
Now the idea is that some costs do not vary with quantity produced. So T can be broken into two parts:
$T = F + V(Q) \implies V(Q) = T  F.$
Now when we look at your specific example $T = 8 + 2Q^2$, the only part of that formula that does not vary as the quantity produced varies is 8.
$So\ F = 8.$ This is conceptually obvious: the only thing in the formula not dependent on Q is 8.
$So\ T = F + V(Q) \implies V(Q) = T  F \implies V(Q) = T  8 = 8 + 2Q^2  8 = 2Q^2.$ Make sense?
OK. If I told you that 5000 units were produced at a total cost of \$200,000, what would you say the average cost is?
$\dfrac{\$200,000}{5000\ units} = \$40\ per \ unit.$ This is fourth grade stuff. Not hard at all.
In English, average cost per unit equals total cost divided by quantity of units produced. In our notation, $A = \dfrac{T}{Q}.$
Putting that into your example, we get:
$A = \dfrac{8 + 2Q^2}{Q} = \dfrac{8}{Q} + \dfrac{2Q^2}{Q} = \dfrac{8}{Q} + \dfrac{2Q * Q}{Q} = \dfrac{8}{Q} + 2Q.$ First year algebra.
OK. Now we get to marginal cost, where the economists use differential calculus. (They should say they use it as an approximation, but they don't.)
Marginal cost is the cost of producing one extra unit. So how do we figure out what that cost is. Well we figure out what is the total cost of producing Q units (we have a formula for that, remember). We figure out using the same formula what is the cost of one more unit. The cost of producing that specific unit is simply the difference in the two costs. So let's work it out for this specific example.
Total cost to produce Q units = $8 + 2Q^2.$
Total cost to produce Q + 1 units = $8 + 2(Q + 1)^2 = 8 + 2(Q^2 + 2Q + 1) = 8 + 2Q^2 + 4Q + 2 = 4Q + 10 + 2Q^2.$
Difference in cost = $(4Q + 10 + 2Q^2)  (8 + 2Q^2) = 4Q + 10  8 + 2Q^2  2Q^2 = 4Q + 2.$
Now if Q is large, then $4Q + 2 \approx 4Q.$ After all, the difference between 2 million dollars and 2 million and 2 dollars is just not significant.
Now the first derivative of $8 + 2Q^2\ is\ 4Q.$
In calculus, you learn how to compute the derivative for different kinds of function. But it looks as though you have been given a formula to memorize for a particular kind of function.
$g(x) = ax^n \implies g'(x) = nax^{(n  1)}.$
When you take calculus, this formula will be proved. But if you are not expected to know calculus, just memorize the formula.