Answer:
a.) C(q) = -(1/4)*q^3 + 3q^2 - 12q + OH b.) $170
Step-by-step explanation:
(a) Marginal cost is defined as the decrease or increase in total production cost if output is increased by one more unit. Mathematically:
Marginal cost (MC) = change in total cost/change in quantity
Therefore, to derive the equation for total production cost, we need to integrate the equation of marginal cost with respect to quantity. Thus:
Total cost (C) = Integral [3(q-4)^2] dq = -(1/4)*(q-4)^3 + k
where k is a constant.
The overhead (OH) = C(0) = -(1/4)*(0-4)^3 + k = -16 + k
C(q) = -(1/4)*(q^3 - 12q^2 + 48q - 64) + k = -(1/4)*q^3 + 3q^2 - 12q -16 + k
Thus:
C(q) = -(1/4)*q^3 + 3q^2 - 12q + OH
(b) C(14) = -(1/4)*14^3 + 3*14^2 - 12*14 + 436 = -686 + 588 - 168 + 436 = $170
C(x) = 200 - 7x + 0.345x^2
Domain is the set of x-values (i.e. units produced) that are feasible. This is all the positive integer values + 0, in case that you only consider that can produce whole units.
Range is the set of possible results for c(x), i.e. possible costs.
You can derive this from the fact that c(x) is a parabole and you can draw it, for which you can find the vertex of the parabola, the roots, the y-intercept, the shape (it open upwards given that the cofficient of x^2 is positive). Also limit the costs to be positive.
You can substitute some values for x to help you, for example:
x y
0 200
1 200 -7 +0.345 = 193.345
2 200 - 14 + .345 (4) = 187.38
3 200 - 21 + .345(9) = 182.105
4 200 - 28 + .345(16) = 177.52
5 200 - 35 + 0.345(25) = 173.625
6 200 - 42 + 0.345(36) = 170.42
10 200 - 70 + 0.345(100) =164.5
11 200 - 77 + 0.345(121) = 164.745
The functions does not have real roots, then the costs never decrease to 0.
The function starts at c(x) = 200, decreases until the vertex, (x =10, c=164.5) and starts to increase.
Then the range goes to 164.5 to infinity, limited to the solutcion for x = positive integers.
Answer:
Step-by-step explanation:
Let's translate what you've written in words to an expression with mathematical symbols.
To solve this, you need to multiply both the numerator and denominator with the conjugate of the denominator.
The graph would look something like this
