Introduction
In this book, the concept of the derivative of a function has been introduced,
and its application in economics has been described. However, the primary use
of the derivative in economic analysis is related to the process of optimization.
Optimization is defined to be the process of determining the local or relative
maximum or minimum of a function.
In this chapter, the process of determining and classifying the relative or
local extrema of a given function is described from a mathematical perspective
by appealing to the local properties of the function near the extrema. The
application of this theory to a range of functions that arise in economics is
described in some detail together with an interpretation of the results. Optimization
is important and useful for solving a range of problems in micro and
macro economics. For example, in the theory of production, the firm wishes to
maximize the output. In microeconomics, a business wishes to maximize profit.
In macroeconomics, a government may wish to maximize revenue from taxation.
The determination of the maxima and minima of a function also provides
invaluable information for the purpose of sketching its graph.
137
138 Elements of Mathematics for Economics and Finance
7.2 Local Properties of Functions
In this section, some local properties of functions are introduced that will be
useful in identifying and characterising the local maxima and minima of a given
function.
7.2.1 Increasing and Decreasing Functions
A function f(x) is said to be increasing on the domain a ≤ x ≤ b if, for any
two points x1, x2, where a ≤ x1 < x2 ≤ b, then f(x1) < f(x2) (see Fig. 7.1(a)). That is, f increases as x increases. A function f(x) is said to be decreasing on the domain a ≤ x ≤ b if, for any two points x1, x2, where a ≤ x1 < x2 ≤ b, then f(x1) > f(x2) (see Fig. 7.1 (b)).
Since the first derivative of a function measures the slope of a function, a
function that is increasing on some domain is characterised by a positive first
derivative. That is, f(x) increases as x takes increasing values in the domain.
More precisely, if f(x) > 0 for all x belonging to some domain a ≤ x ≤ b, then
the function f is said to be increasing for values of x satisfying a ≤ x ≤ b.
Similarly, a function that is decreasing over some domain is characterised by a
negative first derivative. More precisely, if f(x) < 0 for all x belonging to some domain a ≤ x ≤ b, then the function f is said to be decreasing for values of x satisfying a ≤ x ≤ b. For example, the function f(x) = x2 (see Fig. 7.2) is a decreasing function for x < 0 since f(x) = 2x < 0 for x < 0 and an increasing function for x > 0 since f(x) > 0 for x > 0. The function f(x) = 4x − x2 (see
Fig. 7.3) is an increasing function for −1 ≤ x ≤ 2 since f(x) = 4 − 2x > 0
for −1 ≤ x ≤ 2 and a decreasing function for 2 ≤ x ≤ 5 since f(x) < 0 for
2 ≤ x ≤ 5.
7.2.2 Concave and Convex Functions
Consider a function f(x) defined on some domain. If the tangents to the graph
of this function at each point on this domain are such that the graph lies above
them, then the function is said to be convex on the domain. If the tangents to
the graph of this function at each point on this domain are such that the curve
lies below them, then the function is said to be convex on the domain. These
two situations are shown in Fig. 7.4. In the case of the convex function shown
in Fig. 7.4(a), we observe that the slope of the function increases as one moves
from the point x1 to the point x2. In this particular example, the slope of the
function is negative at x = x1 and gradually increases to take a positive value

7. Maxima and Minima 139
   x
   y
   x1 (a) x2 x
   y
   x x 1 2 (b)
   Figure 7.1 Examples of graphs of (a) an increasing function; (b) a decreasing
   function.
   at x = x2. Thus, a function that is convex on a domain is characterised by
   the condition f(x) > 0 on the domain. Similarly, a function that is concave
   on a domain is characterised by the condition f(x) < 0 on the domain (see Fig.7.4(b)). For example, the function f(x) = x2 (see Fig. 7.2) is convex on the domain −2 ≤ x ≤ 2. In fact, it is convex on any domain a ≤ x ≤ b since f(x) = 2 > 0.
   The function f(x) = 4x−x2 (see Fig. 7.3) is concave on the domain −1 ≤ x ≤ 5.
   In fact, it is concave on any domain a ≤ x ≤ b since f(x) = −2 < 0.
   7.3 Local or Relative Extrema
   A function of x possesses a local maximum or minimum at x = a if the function
   is neither increasing nor decreasing at x = a. That is, the rate of change of y
   relative to x is 0 when x = a. A local or relative extremum of a function is a
   point at which the function attains a local maximum or minimum. This means
   that the tangent to the curve y = f(x) is ‘horizontal’ at a local or relative
   extremum and therefore has zero slope. Equivalently, since the slope is given
   by the first derivative of the function, that derivative must be zero at x = a. A
   point where f(x) = 0 is known as a critical point or value. It is also known
   as a stationary point.
   So the stationary or critical points of a function f(x) = 0 are the values of
   x for which f(x) = 0. It remains to classify them as maxima or minima. This
   is done by calculating the second derivative of the function and evaluating it
   140 Elements of Mathematics for Economics and Finance
   x
   -2 -1 0 1 2
   1
   2
   3
   y 4
   Figure 7.2 The graph of the function f(x) = x2.
   -1 0 1 2 3 4 5
   -5
   -3
   -1
   1
   3
   y
   x
   Figure 7.3 The graph of the function f(x) = 4x − x2.
8. Maxima and Minima 141
   x
   y
   x x 1 2 (a) x
   y
   x1 x2
   (b)
   Figure 7.4 Examples of graphs of (a) a convex function; (b) a concave
   function.
   at x = a.
   To distinguish between a relative maximum and a relative minimum, it is
   necessary to inspect the behaviour of the second derivative and, in particular,
   to determine the sign of f(a) where f(a) = 0. For points just to the left of a
   local maxima at x = a, the slope of the tangent is positive, and for points just
   to the right, the slope of the tangent is negative. So in the neighbourhood of
   a local maxima, the first derivative of f(x) is a decreasing function of x, i.e.,
   f(x) < 0 and, in particular, f(a) < 0. Therefore, if f(a) < 0, which means that the function is concave and the curve lies below the tangent at x = a, then the function has a local maximum at x = a. For points just to the left of a local minima at x = a, the slope of the tangent is negative, and for points just to the right, the slope of the tangent is positive. So in the neighbourhood of a local minima, the first derivative of f(x) is an increasing function of x (i.e., f(x) > 0) and, in particular, f(a) > 0. Therefore, if f(a) > 0, which means
   that the function is convex and the curve lies above the tangent at x = a, then
   the function has a local minimum at x = a.
   We now summarize the steps involved in finding and classifying the stationary
   points of a function f(x):
   Second Derivative Test
   Step 1.
   Solve the equation
   f
   (x) = 0
   to find the stationary point(s).
   142 Elements of Mathematics for Economics and Finance
   Step 2.
   Suppose x = a gives a stationary point (i.e., f(a) = 0).
   If f(a) > 0 then the function has a local minimum at x = a.
   If f(a) < 0 then the function has a local maximum at x = a.
   If f(a) = 0 then the test is inconclusive.
   A knowledge of the stationary points of a function is essential when sketching
   the graph of a nonlinear function since it provides information about its
   general shape. The graph of a function can be sketched using a similar process
   to that used to determine and classify the stationary points of a function. Once
   the stationary points of a function have been determined and classified, the
   graph of the function can be sketched by drawing a smooth curve through
   these points. A more accurate representation of the graph may be obtained by
   evaluating the function at a greater number of points and drawing a smooth
   curve through them.
   Example 7.1
   Find and classify the stationary points of the following functions:
9. f(x) = x2 − 4x + 5,
10. f(x) = 2×3 + 3×2 − 12x + 4.
    Solution.
11. We need to calculate the first and second order derivatives of f(x) = x2 −
    4x + 5.
    f
    (x) = 2x − 4
    f
    (x) = 2
    Step 1. The stationary points are the solutions of the equation
    f
    (x) = 0,
    i.e.,
    2x −4 = 0.
    Therefore x = 2 is a stationary point.
    Step 2. To classify this point, we need to evaluate f(2). In this case,
    f(2) = 2 > 0 so the function has a minimum at x = 2. The graph of this
    function is shown in Fig. 7.5.
12. Maxima and Minima 143
    -1 0 1 2 3 4 5
    2
    4
    6
    8
    10
    y
    x
    Figure 7.5 The graph of the function f(x) = x2−4x+5 plotted for values
    of x lying between −1 and 5.
13. In this example, we have f(x) = 2×3 + 3×2 − 12x + 4
    f
    (x) = 6×2 + 6x − 12
    f
    (x) = 12x + 6
    Step 1. The stationary points are the solutions of the equation f(x) = 0,
    i.e.,
    6(x2 + x − 2) = 0
    6(x + 2)(x − 1) = 0
    Therefore, the stationary points are x = −2 and x = 1.
    Step 2. To classify these points, we need to evaluate f(x) at x = −2 and
    x = 1. Now,
    f
    (−2) = −24 + 6 = −18 < 0, and so the function has a maximum at x = −2, and f (1) = 12 + 6 = 18 > 0,
    so the function has a minimum at x = 1. The graph of this function is
    shown in Fig. 7.6.
    144 Elements of Mathematics for Economics and Finance
    -4 -3 -2 -1 0 1 2 3
    -20
    -10
    10
    20
    30
    40
    y
    x
    Figure 7.6 The graph of the function f(x) = 2×3 +3×2 −12x+4 plotted
    for values of x lying between −4 and 3.
    7.4 Global or Absolute Extrema
    The functions we have investigated so far have either possessed a single stationary
    point (see Figs. 7.2 and 7.3) or two stationary points (see Fig. 7.6). In
    general, however, we may encounter functions that possess several local extrema
    of the same type. For example, the function f(x) = 2
    5×5+ 3
    4×4−8×3−3×2+12x
    (see Fig. 7.7) has local maxima at x = −4 and x = 1/2 and local minima at
    x = −1 and x = 3. The higher of the two local maxima occurs at x = −4.
    However, the largest value of f(x) for values of x lying between −4 and 3 occurs
    at the end point x = 3. We say that this point is a global or absolute
    maximum. The lower of the two local minima occurs at x = 3. However,
    the smallest value of f(x) for values of x lying between −4 and 3 occurs at
    the other end point x = −4. We say that this point is a global or absolute
    minimum. Note that the absolute maximum and absolute minimum values of
    this function are not stationary points since the slope of the function is not
    zero at either x = −4 or x = 3. This example demonstrates that the absolute
    maximum or absolute minimum values of a function defined in a given interval
    may not occur at local extrema.
14. Maxima and Minima 145
    x
    -5 -4 -3 -2 -1 0 1 2 3 4
    -50
    50
    150
    250 y
    Figure 7.7 The graph of the function f(x) = 2
    5×5+ 3
    4×4−8×3−3×2+12x
    plotted for values of x lying between −5 and 4.
    There is no method for determining global extrema other than to evaluate
    the function at all local extrema and the end points and to determine from
    these calculations the values of x that generate the global extrema. For most of
    the examples we encounter in economics, the local extrema will coincide with
    the global extrema.
    7.5 Points of Inflection
    The local extrema of a function f(x) have been characterised by the solutions
    x = a of the equation f(x) = 0 and classified as being local maxima or
    local minima depending on whether f(a) < 0 or f(a) > 0, respectively.
    So far, we have not asked what happens if f(a) = 0. In this situation, the
    second derivative test is inconclusive and the stationary point x = a is either
    a local maxima, a local minima, or a point of inflection. At a stationary point
    of inflection, the function is neither convex nor concave. The function crosses
    its tangent at this point and changes from concave to convex or vice versa.
    146 Elements of Mathematics for Economics and Finance
    x
    -2 -1 0 1 2
    -7
    -5
    -3
    -1
    1
    3
    5
    7
    y
    Figure 7.8 The graph of the function f(x) = x3 plotted for values of x
    lying between −2 and 2.
    For example, the function f(x) = x3 (see Fig. 7.8) has a stationary point of
    inflection at x = 0. For this function, we have f(0) = f(0) = 0. In addition,
    the function is increasing for all values of x, convex for x < 0 and concave for x > 0. The function changes from being convex to concave at the point x = 0.
    It is possible to have a point of inflection that is not a stationary point.
    For example, the function f(x) = x3 − 3×2 + 2x (see Fig. 7.9) has a point of
    inflection at x = 1. At this point, we have f(1) = 0 but f(1) = −1 = 0.
    7.6 Optimization of Production Functions
    Production functions were introduced in Chapter 5. Production depends on a
    number of factors including capital and labour. However, in the short run we can
    assume that a firm’s production depends solely on labour with all other factors
    of production, including capital, constant. We can express this symbolically by
    writing
    Q = Q(L).
15. Maxima and Minima 147
    -1 0 1 2 3
    -7
    -5
    -3
    -1
    1
    3
    5
    y
    x
    Figure 7.9 The graph of the function f(x) = x3 − 3×2 + 2x plotted for
    values of x lying between −1 and 3.
    The marginal product of labour, MPL, is the derivative of the output with
    respect to labour and is defined by
    MPL = dQ
    dL
    = Q
    (L). (7.1)
    Under the assumption that production depends on labour alone, it is possible
    to calculate the size of the workforce that maximizes production. The following
    example illustrates this process.
    Example 7.2
    A firm’s short run production function is given by
    Q = 6L2 − 0.2L3
    where L denotes the number of workers.
16. Find the size of the workforce that maximizes output and hence sketch a
    graph of this production function.
17. Find the size of the workforce that maximizes the average product of
    labour. Calculate MPL and APL at this value of L. What do you observe?
    148 Elements of Mathematics for Economics and Finance
    Solution.
18. To solve the first part of this example, it is necessary to determine and
    classify the stationary points of the production function.
    Step 1. At a stationary point of the production function
    dQ
    dL
    = 12L − 0.6L2 = 0.
    Therefore
    L(12 − 0.6L) = 0
    and so L = 0 or L = 12/0.6 = 20.
    Step 2. It is obvious on economic grounds that L = 0 gives the minimum
    Q = 0. We can, of course, check this by differentiating a second time to get
    d2Q
    dL2 = 12− 1.2L.
    When L = 0,
    d2Q
    dL2 = 12 > 0,
    which confirms that L = 0 gives a minimum for Q.
    When L = 20,
    d2Q
    dL2 = 12− 24 = −12 < 0,
    thus L = 20 gives a maximum for Q.
    The firm should therefore employ 20 workers to achieve a maximum output
    Q = 6(20)2 − 0.2(20)3 = 800.
    The graph of this production function is sketched in Fig. 7.10.
19. To solve the second part of the problem, it is necessary to determine and
    classify the stationary point of the average product of labour, APL,
    which is defined by
    APL = Q
    L
    . (7.2)
    This is sometimes called labour productivity since it measures the average
    output per worker. In this example,
    APL =
    6L2 − 0.2L3
    L
    = 6L − 0.2L2.
    Step 1. At a stationary point
    d(APL)
    dL
    = 0,
20. Maxima and Minima 149
    L
    5 10 15 20 25 30 35
    0
    200
    400
    600
    800
    Q
    Figure 7.10 The graph of the production function Q = 6L2 − 0.2L3.
    i.e.,
    6 − 0.4L = 0,
    and therefore L = 6/0.4 = 15.
    Step 2. To classify this stationary point, we differentiate a second time to
    get
    d2(APL)
    dL2 = −0.4 < 0
    which shows that it is a maximum. The labour productivity is therefore
    greatest when the firm employs 15 workers. The corresponding labour productivity
    is
    APL = 6(15) − 0.2(15)2 = 45.
    So the largest number of goods produced per worker is 45.
    To find an expression for MPL, we need to differentiate Q with respect to
    L, which we have already done in the first part of the problem. We have
    MPL = dQ
    dL
    = 12L − 0.6L2.
    When L = 15,
    MPL = 12(15) − 0.6(15)2 = 45.
    We observe that at L = 15, the values of MPL and APL are equal.
    150 Elements of Mathematics for Economics and Finance
    In this example, we have discovered that:
    At the point of maximum average product of labour,
    marginal product of labour = average product of labour,
    i.e., MPL = APL.
    In fact, this result holds for any production function Q = Q(L) as we
    shall demonstrate. If we differentiate the expression (7.2) defining the average
    product of labour using the quotient rule, we obtain
    d(APL)
    dL
    = d(Q/L)

dL

LdQ
dL
− QdL
dL
L2
= Q(L) − Q(L)/L
L
= MPL − APL
L
. (7.3)
At a stationary point for the average product of labour, we have
d(APL)
dL
= 0. (7.4)
This means that MPL = APL, as required. This result shows that at a stationary
point of the average product of labour, the marginal product of labour
is equal to the marginal product of labour. The analysis has shown that this
result is true for any function APL and is not restricted to certain choices.
At a stationary point of the average product of labour, we can obtain a
simple expression for the second derivative of APL with respect to L as follows:
d2(APL)
dL2 = d
dL

MPL − APL
L

L

d(MPL)
dL
− d(APL)
dL

− (MPL − APL)dL
dL
L2 .
At a stationary point, we know MPL = APL and also d(APL)/dL = 0. Therefore,
d2(APL)
dL2 =
1
L
d(MPL)

dL

1
L
d2Q
dL2 ,
since MPL = dQ/dL. So at a stationary point of the average product of labour

7. Maxima and Minima 151
   d2(APL)
   dL2 = Q(L)
   L
   .
   7.7 Optimization of Profit Functions
   We turn our attention to the problem of determining the maximum profit for
   a given firm. The profit function, π, which is the difference between the total
   revenue and total cost functions, is expressed as a function of the output Q.
   The function is then optimized with respect to Q.
   Example 7.3
   Maximize the profit for a firm, given that its total revenue function is given
   by TR = 4,000Q − 33Q2 and its total cost function by TC = 2Q3 − 3Q2 +
   400Q + 5,000, assuming Q > 0.
   Solution. The profit function is given by
   π = TR − TC
   = 4,000Q − 33Q2 − (2Q3 − 3Q2 + 400Q + 5,000)
   = −2Q3 − 30Q2 + 3,600Q − 5,000
   Step 1. At a stationary point of the profit function,
   dπ
   dQ
   = 0.
   Now
   dπ
   dQ
   = −6Q2 − 60Q + 3,600
   = −6(Q2 + 10Q − 600)
   = −6(Q + 30)(Q − 20).
   Therefore, the stationary points of the profit function are Q = −30 or Q = 20.
   (As an alternative to factorization, the equation
   dπ
   dQ
   = 0
   can also be solved using the quadratic formula (3.8).)
   152 Elements of Mathematics for Economics and Finance
   Step 2. The stationary point Q = −30 has no economic significance since π
   is only defined for Q > 0. Therefore, we can ignore it. To classify the second
   stationary point, we differentiate a second time
   d2π
   dQ2 = −12Q − 60.
   When Q = 20,
   d2π
   dQ2 = −240 − 60 = −300 < 0,
   which shows that π has a local maximum when Q = 20. Therefore, the profit
   is maximized when Q = 20 and the maximum profit is given by
   π = −2(20)3 − 30(20)2 + 3,600(20) − 5,000
   = −16,000 − 12,000 + 72,000 − 5,000
   = 39,000.
   Example 7.4
   The demand equation for a good is
   P + Q = 30
   and the total cost function is
   TC =
   1
   2Q2 + 6Q + 7.
8. Find the level of output that maximizes total revenue.
9. Find the level of output that maximizes profit. Calculate MR and MC at
   this value of Q. What do you observe?
   Solution.
10. The total revenue function is defined by TR = P ×Q. Now P = 30−Q
    by rearranging the demand equation. Therefore,
    TR = (30− Q)Q = 30Q − Q2.
    Therefore,
    d(TR)
    dQ
    = 30 − 2Q.
    Step 1. At a stationary point of the total revenue function
    d(TR)
    dQ
    = 0,
11. Maxima and Minima 153
    so
    30 − 2Q = 0.
    Therefore Q = 15.
    Step 2. To classify this point, we differentiate a second time to get
    d2(TR)
    dQ2 = −2 < 0,
    so TR has a local maximum when Q = 15.
12. The profit function is defined by
    π = TR − TC
    = (30Q − Q2) − (
    1
    2Q2 + 6Q + 7)
    = −3
    2Q2 + 24Q − 7.
    Therefore,
    dπ
    dQ
    = −3Q + 24.
    Step 1. At a stationary point of the profit function
    dπ
    dQ
    = 0,
    so
    −3Q + 24 = 0,
    which has the solution Q = 8.
    Step 2. To classify the stationary point, we differentiate the profit function
    a second time to get
    d2π
    dQ2 = −3 < 0,
    so π has a local maximum at Q = 8. Now
    MR = d(TR)
    dQ
    = 30− 2Q,
    and
    MC = d(TC)
    dQ
    = Q + 6.
    Therefore, when Q = 8, then MR = 14 and MC = 14. So then
    At the value of Q that maximizes profit,
    marginal revenue = marginal cost
    154 Elements of Mathematics for Economics and Finance
    This result is true for any profit function irrespective of the market conditions
    under which the firm operates since at a stationary point for the profit
    function, we have
    dπ
    dQ
    = d(TR)
    dQ
    − d(TC)
    dQ
    = MR−MC = 0.
    Therefore, MR = MC at a stationary point for the profit function.
    7.8 Other Examples
    Example 7.5
    The cost of building an office block, x floors high, comprises three components:
13. £18 million for the land,
14. £200,000 per floor,
15. specialized costs of £20,000x per floor. (Thus if there are to be 4 floors,
    the specialized cost per floor will be £80,000.)
    How many floors should the block contain if the average cost per floor is to be
    minimized?
    Solution. First of all, we need to derive an expression for the total cost of
    construction of the office block. Suppose that the building has x floors. Then the
    £18 million is a fixed cost because it is independent of the number of floors.
    The total cost involved in the second component is £200,000x. In addition,
    there are specialized costs of £20,000x per floor. So if there are x floors, the
    specialized costs will be
    (20,000x)x = 20,000×2.
    The total cost of construction in terms of monetary units of £1,000 is therefore
    TC = 18,000 + 200x + 20×2.
    The average cost per floor, AC, is formed by dividing the total cost by the
    number of floors, i.e.,
    AC = TC

x

18,000 + 200x + 20×2

x

18,000
x

• 200 + 20x

1. Maxima and Minima 155
   Step 1. At a stationary point
   d(AC)
   dx
   = 0.
   Now
   d(AC)
   dx
   = −18,000x
   −2 + 20.
   So we need to solve the equation
   −18,000x
   −2 + 20 = 0.
   Therefore,
   20 = 18,000x
   −2 =
   1,800
   x2
   1 =
   900
   x2 (Divide both sides by 20.)
   x2 =
   900
   x2 x2 (Multiply both sides by x2.)
   x2 = 900
   Therefore x2 = 900 and so x = ±
   √
   900 = ±30.
   Step 2. To confirm that x = 30 yields a minimum, we need to differentiate a
   second time.
   d2(AC)
   dx2 = 36,000x
   −3
   So obviously when x = 30,
   d2(AC)
   dx2 > 0.
   Thus x = 30 gives a minimum for AC.
   Therefore an office block 30 floors high produces the lowest average cost per
   floor.
   Example 7.6
   The supply and demand equations of a good are
   P = Qs + 8,
   P = −3Qd + 80,
   respectively. The government decides to impose a tax, et, per unit of good.
   Find the value of t that maximizes the government’s total tax revenue on the
   assumption that equilibrium conditions prevail in the market.
   156 Elements of Mathematics for Economics and Finance
   Solution. To account for the imposition of tax, we replace P by P − t in the
   supply equation. This is because the price that the supplier actually receives is
   the price P that the consumer pays, less the tax t deducted by the government.
   The new supply equation is then
   P − t = Qs + 8,
   so that
   P = Qs +8+t.
   In equilibrium,
   Qs = Qd.
   If this common value is denoted by Q then the demand and supply equations
   are
   P = −3Q + 80,
   P = Q+8+t.
   Hence,
   Q+8+t = −3Q + 80.
   Therefore
   4Q = 72− t,
   and so
   Q = 18− 1
   4t.
   Now if the number of goods sold is Q and the government raises t per good,
   then the total tax revenue is given by
   T = tQ
   = t(18 − 1
   4t)
   = 18t − 1
   4t2.
   This is the function we wish to maximize.
   Step 1. At a stationary point,
   dT
   dt
   = 0,
   so
   dT
   dt
   = 18 − 1
   2t = 0,
   which has the solution t = 36.
   Step 2. To classify the stationary point, we differentiate a second time to get
   d2T
   dt2 = −1
   2 < 0,
   which confirms that it is a maximum. Hence the government should impose a
   tax of e36 on each good.
2. Maxima and Minima 157
   EXERCISES
   7.1. For the following functions, determine the stationary point(s) and
   classify them. Use this information to sketch graphs of these functions.
   a) f(x) = 2×2 − x + 6.
   b) f(x) = x2 − 4x + 3.
   c) f(x) = x3 − 3x + 3.
   d) f(x) = 1 − 9x − 6×2 − x3.
   7.2. The demand equation for a good is given by
   P + 4Q = 96,
   and the total cost function TC is
   TC = Q3 − 13Q2 + 48Q + 17.
   a) Find the level of output that maximizes total revenue.
   b) Find the maximum profit and the level of output for which it is
   achieved.
   c) Sketch the graph of profit against Q, for Q ≥ 0.
   7.3. The demand equation for a good is given by
   P + 2Q = 20,
   and the total cost function TC is
   TC = Q3 − 8Q2 + 20Q + 2.
   a) Find the level of output that maximizes total revenue.
   b) Find the maximum profit and the level of output for which it is
   achieved. Verify that, for this value of Q, MR = MC.
   7.4. The prevailing market price for a good is 30. The total cost function
   is
   TC = 100 + 44Q − 5Q2 +
   1
   2Q3.
   What is the level of output that maximizes the profit?
   158 Elements of Mathematics for Economics and Finance
   7.5. Find the first and second order derivatives, with respect to L, of the
   short run production function
   Q = 15L2 − 2L3.
   Hence, determine and classify the stationary points of this function.
   7.6. A firm’s short run production function is given by
   Q = 12L2 − 1
   2L3,
   where L denotes the number of workers.
   a) Find the size of the workforce that maximizes output and hence
   sketch a graph of this production function.
   b) Find the size of the workforce that maximizes the average product
   of labour, APL. Calculate MPL and APL at this value of L.
   What do you observe?
   7.7. The cost of building an office block, x floors high, is made up of three
   components:
   a) $20.16 million for the land,
   b) $175,000 per floor,
   c) specialized costs of $35,000x per floor.