Exam (elaborations)
DSC1520 Assignment 03 Semester 1 2023 (860914)
- Institution
- University Of South Africa (Unisa)
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Quantitative Modelling 1DSC1520
ASSIGNEMNT 03 2023 (860914)
SEMESTER 1 2023
QUESTION 1
Bongi supplies trays of fresh sandwiches to offices daily. Her daily fixed cost amount
to R844, while her variable cost is R27 per tray. Bongi's total cost and marginal cost
functions (in terms of the number of trays supplied, Q are given by
a. TC=844+27Q; MC=27
b. TC=844+27Q; MC=27Q
c. TC=27; MC= 844+27Q
d. TC=27Q; MC=844+27Q
TC = FC +VC = 844 + 27Q
MC(q) = d(TC)/d(q) = 27
QUESTION 2
Suppose we have the demand function given as
p=72−7,5q
where q is the number of units to be produced and sold. Determine the marginal revenue
after 3 units have been sold.
a. R1
b. R149
c. R50
d. R27
, The total revenue function is given by TR = p*q, where p is the price and q is the quantity
sold. Substituting the demand function into this equation, we get:
TR(q) = (72 - 7.5q) * q
TR(q) = 72q - 7.5q^2
The marginal revenue is the derivative of the total revenue with respect to the quantity
sold:
MR(q) = d(TR)/d(q)
MR(q) = 72 - 15q
To find the marginal revenue after 3 units have been sold, we substitute q = 3 into the
marginal revenue equation:
MR(3) = 72 - 15(3) = 72 - 45 = 27
Therefore, the marginal revenue after 3 units have been sold is R27.
QUESTION 3
Determine the intervals along which the given function f(n) increases or decreases, where
f(n)=30n2−n3.
a. The function f(n) decreases on the interval (0;20), while it increases from the
interval (−∞;0)(−∞;0) and (20;∞)(20;∞).
b. The function f(n) increases on the interval (0;20)(0;20) and (−∞;0)(−∞;0), while it
decreases from the interval (20;∞)(20;∞).
c. The function f(n) increases on the interval (0;20)(0;20) and (20;∞)(20;∞), while it
decreases from the interval (−∞;0)(−∞;0).
d. The function f(n) increases on the interval (0;20)(0;20), while it decreases from the
interval (−∞;0)(−∞;0) and (20;∞)(20;∞).
To determine the intervals of increase and decrease for the function f(n) = 30n^2 - n^3,
we need to find its derivative and analyze its sign.
Taking the derivative of f(n), we get:
f'(n) = 60n - 3n^2
Setting f'(n) equal to zero, we get:
60n - 3n^2 = 0
3n(20 - n) = 0
n = 0 or n = 20
Now we can check the sign of the derivative in the intervals (-∞, 0), (0, 20), and (20,
∞):
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