BUSI1074 QUANTITATIVE METHODS 1b 2022-2023 Worksheet 1
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QUANTITATIVE METHODS 1b – BUSI1074
A.Y. 2022-2023
Worksheet 1 - Solutions
Exercise 1
Find derivatives of the following functions:
• y = ln(4X3 (−16 + 4X)5)
Answer:
dy d d
dX dX dX
= + = = =
5
• y = a
Answer
y = ax5 = eln(a"5) = ex5 ln(a) , therefore
dy
dX
or alternatively, we can solve by taking the logarithm of function y:
lny = X5lna, differentiate both sides and get: l. ℎ .S.: = , T. ℎ .S.: = 5X4lna. Since l. ℎ .S. = T. ℎ .S., we have:
dy
dX
Exercise 2
Use implicit differentiation to find expression dy /dx for function lnX + ey − 4X = 4.
Answer:
Let now function z be Z = 4 = lnX + ey − 4X. By the implicit differentiation theorem it is easy to derive
#$ 1
= − = − = − =
#y
Indeed, dZ = d4 = dX + dy = 0 and thus M − 4N dX + eydy = 0, which yields
= − =
Exercise 3
Find the equation of a quadratic function y = ax2 +bx + c (where a, b and c are constants to be determined) whose graph is tangent at x = 2 to the line with slope 4, tangent at x = -4 to the line with slope -8 and tangent to the line y = 6.
Answer:
At each point of graph y, the slope of the tangent curve is = 2a + b.
As y is tangent at x = 2 to the line with slope 4, it must be true that 4a +b = 4 at x = 2 . As y is tangent at x = -4 to the line with slope -8, then - 8a +b = -8 at x = -4.
Hence, we can find a = 1 and b = 0 , thus y = x2 + c . Furthermore, we know that the graph is tangent to the horizontal line y = 6. We can find the values of x such that the derivative of function y at those points is equal to zero (remember that the slope of horizontal lines is zero). This value is x = 0, which gives c = 6 . Finally, the quadratic function is y = x2 + 6 .
Exercise 4
Consider the demand function P = . Find the price elasticity of demand (PED) at point Q = 18 and interpret the result.
Answer:
The PED, or simply E, is defined as
E = − ∙ = −
where = (100 − 2Q)-1/2 ∙ (−2) = − . Thus:
E = − ∙ = T100 − 2Q ∙
At Q = 18 (and corresponding P = 8), we have:
E = − dQ P 8 64
Interpretation: when the price changes by 1% (either an increase or a decrease) the quantity demanded changes (either a decrease or an increase) by 3.55%.
Here, we used the fact that dQ/dP is 1/(dP/dQ). This is not because of changing the order of denominator and nominator. Remember that dP and dQ do not have independent meanings. Rather, this comes from the implicit differentiation rule. Indeed, another way of solving this question is to use implicit differentiation to P = . Thinking of Q as an output and P
as an input, let us differentiate it w.r.t. P. This will give 1= ( 100-2Q)^{- 1/2}*(-2)*dQ/dP. Rearranging the term, we have the same answer.
Note that: Obviously, the elasticityper sè M N is negative, because the demand curve slopes downwards, but it is conventional to avoid the negative sign by changing it into positive, so getting a general index.
Formula E = − is equivalent to formula E = VV, where the later takes the absolute value of the negative value , which still yields a positive value. Alternatively, you might find in some books the negative value for PED (or E), for instance the formula E = , but in order to judge whether the demand is elastic or inelastic still the absolute value should be considered. It is only a matter of convention, nothing changes.
For instance, the textbook Quantitative Modules presents the sign ‘- ’, while the textbook Mathematicsfor Economics and Businessjust gives you the negative valuefor E, but then it is recommended to consider the absolute value to determine whether the demand is elastic or not. So, in the answer below (Ex. 5), we use again the sign ‘-’ and judge directly about elastic/inelastic demand, which is of course equivalent to using the absolute value.
Exercise 5
Consider the demand function P = 30 - Q and the total cost function TC = Q2 + 6Q + 7.
a) Find an expression for TR in terms of Q. TR = PQ = 30Q – Q2
b) Find the marginal cost function.
The marginal cost function must be MC = dTC/dQ = Q + 6
c) Find the average cost function. AC = TC/Q = (1/2)Q + 6 + 7/Q
d) Find the level of output that maximizes profits. The expression for profits, in terms of Q, is
π = TR – TC = Q(30 - Q) - 1/2 Q2 - 6Q - 7 = (- 3/2)Q2 + 24Q - 7 . The stationary point is found at = 0 .
= -3Q+ 24 = 0 has solution Q = 8. To classify this point, we differentiate a second time: = -3 . This derivative is negative, so Q* = 8 is a max and the profit is π* = (- 3/2)82 + 24 *8 - 7 = 89 .
e) Verify that, at this value of Q, MR = MC.
MR = d(TR)/dQ = 30 - 2Q
MR(8) = 30-2*8 = 14
MC(8) = 8 + 6 = 14
f) Demonstrate that at the maximum profit it must hold < .
At the maximum, by definition, it must be true that = , that is MR = MC from the
d冗 d(TR) d(TC)
dQ dQ dQ
derivative must be negative (second derivative test), that is d(d)Q(2)2(冗) < 0, implying < .
Exercise 6
Find and classify the stationary points of function f(x) = 2x3 − 4x2 + 2x + 1 .
Answer:
First, solve the first order condition:
f ′ (x) = 6x2 − 8x + 2 = 0, which gives critical points x = 1 and x = , with f(1) = 1 and f(1/3) = .
f,,(x) = 12x − 8
f,, (1) = 12 ∗ 1 − 8 = 4 > 0
f,,(1/3) = 12 ∗ − 8 = −4 < 0
Thus point (1,1) is a minimum, whereas point ( , ) is a maximum.
-2.0 - 1.5 - 1.0 -0.5 y 0.5 1.0 1.5 2(x)0
- 10
-20
-30
Note that: Given afunction y = f(x), when we refer to a point -for example point (1,1) - we sometimes use the abbreviation ‘point x=1’, but it is obvious that every point on the graph y = f(x) is by definition defined by two coordinates x and y. In this case, x=1 should not be interpreted as a vertical line at value x=1, because otherwise we have to refer to it as afunction, or as a graph, and therefore we would say ‘function x=1’ or ‘graph x=1’, rather than ‘point
x=1’.
(*) During the exam, there won’t be no confusion: if we want you to find the maximizing x, it will say so. If we want you to find the maximizing x and corresponding y, it will be explicitly said.
2023-01-12