In the previous chapter we differentiated simple functions: powers of x, and polynomials. Now we introduce further techniques of differenti-ation, such as the product, quotient and chain rules, and the rules for differentiating exponential and logarithmic functions. We use differentiation in a variety of economic applications: for example, to find the profit-maximising output for a firm, the growth rate of GDP, the elasticity of demand and the optimum time to sell an asset.
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1. Graph Sketching
The first step towards understanding an economic model is often to draw pictures: graphs of supply and demand functions, utility functions, production functions, cost functions, and so on. In previous chapters we have used a variety of methods, including differentiation, to work out the shape of a function. Here it is useful to summarise them, so that you can use them in the economic applications in this chapter.
1.1. Guidelines for Sketching the Graph of a Function y(x)
• Look at the function y(x) itself:
– to find where it crosses the x- and y-axes (if at all), and whether it is positive or negative elsewhere;
– to see what happens to y when x goes towards ±∞;
– to see whether there are any values of x for which y goes to ±∞.
• Look at the first derivative y0(x) to find the gradient:
– It may be positive for all values of x, so the function is increasing;
or negative for all values of x, so the function is decreasing.
– Otherwise, find the stationary points where y0(x) = 0, and check whether the graph slopes up or down elsewhere.
• Look at the second derivative y00(x) to see which way the function curves:
– Use it to check the types of the stationary points.
– It may be positive for all values of x, so the function is convex;
or negative for all values of x, so the function is concave.
– Otherwise, it tells you whether the gradient is increasing or decreasing in dif-ferent parts of the graph.
Depending on the function, you may not need to do all of these things. Sometimes some of them are too difficult – for example, you may try to find where the function crosses the
97
x-axis, but end up with an equation that you can’t solve. But if you miss out some steps, the others may still give you enough information to sketch the function. The following example illustrates the procedure.
Examples 1.1: Sketch the graph of the function y(x) = 2x3− 3x2− 12x (i) y(0) = 0, so it crosses the y-axis at y = 0
(ii) To find where it crosses the x-axis, we need to solve the equation y(x) = 0:
2x3− 3x2− 12x = 0 x(2x2− 3x − 12) = 0
One solution of this equation is x = 0. To find the other solutions we need to solve the equation 2x2− 3x − 12 = 0. There are no obvious factors, so we use the quadratic formula:
2x2− 3x − 12 = 0 x = 3 ±√
105 4
x ≈ 3.31 or x ≈ −1.81 So the function crosses the x-axis at −1.81, 0, and 3.31.
(iii) What happens to y(x) = 2x3− 3x2− 12x as x becomes infinite?
For a polynomial, it is the term with the highest power of x that determines what happens when x becomes infinite – it dominates all the other terms. In this case:
2x3 → ∞ as x → ∞, so y(x) → ∞ as x → ∞ 2x3 → −∞ as x → −∞, so y(x) → −∞ as x → −∞
(iv) Are there any finite values of x that make y(x) infinite?
No; we can evaluate y(x) for all finite values of x.
(v) The first derivative of the function is:
y0(x) = 6x2− 6x − 12 Stationary points occur where y0(x) = 0:
6x2− 6x − 12 = 0 x2− x − 2 = 0 (x + 1)(x − 2) = 0
x = −1 or x = 2
So there are stationary points at x = −1, y = 7 and x = 2, y = −20 (vi) The second derivative of the function is:
y00(x) = 12x − 6 At (−1, 7), y00(x) = −18 so this is a maximum point.
At (2, −20), y00(x) = 18 so this is a minimum point.
(vii) The second derivative is zero when x = 12.
When x < 12, y00(x) < 0, so the gradient is decreasing.
When x > 12, y00(x) > 0, so the gradient is increasing.
To draw the graph, first put in the points where the function crosses the axes, and the stationary points. Then fill in the curve, using what you know about what happens as x → ±∞, and about where the gradient is increasing and decreasing:
y(x)
1.2. Economic Application: Cost Functions
Examples 1.2: A firm’s total cost of producing output y is C(y) = 2y2+ 3y + 8.
(i) What are its fixed and variable costs?
Fixed costs are costs that must be paid whatever the level of output; variable costs depend on the level of output. If the firm produces no output, its costs are: C(0) = 8.
So fixed costs are F = 8, and variable costs are V C(y) = 2y2+ 3y (ii) What is the marginal cost function?
The marginal cost function is the derivative of the cost function:
M C(y) = C0(y) = 4y + 3 (iii) What is the average cost function?
AC(y) = C(y)
y = 2y + 3 +8 y
(iv) Show that the marginal cost curve passes through the point of minimum average cost.
The derivative of the average cost function is:
dAC
dy = 2 − 8 y2
The average cost function has stationary point(s) where its derivative is zero:
2 − 8
y2 = 0 ⇒ y = 2 (ignoring negative values of y)
Look at the second derivative to see if this is a miminum:
d2AC dy2 = 16
y3
Since this is positive when y = 2 we have found a minimum point. At this point:
y = 2, AC(y) = 11 and M C(y) = 11
So the marginal cost curve passes through the point of minimum average cost.
(v) Sketch the marginal and average cost functions.
We will sketch the functions for positive values of y only, since negative values are of no economic interest. Sketching M C(y) is easy: it is a straight line, of gradient 4, crossing the vertical axis at 3, and (from above) passing throught the point (2,11).
For the average cost function:
AC(y) = C(y)
y = 2y + 3 +8 y
we already know that it has one stationary point, which is a minimum. Also, we have found the second derivative, and it is positive for all positive values of y, so the gradient is always increasing.
We can see by looking at AC(y) that as y → 0, AC(y) → ∞, and that as y → ∞, AC(y) → ∞. The function does not cross either of the axes.
.
Exercises 6.1: Sketching Graphs (1)
(1) Sketch the graphs of the functions:
(a) a(x) = (1 − x)(x − 2)(x − 3) (b) b(x) = x12(x − 6) for x ≥ 0.
(2) A firm has cost function C(y) = y3+ 54. Show that the marginal cost function passes through the point of minimum average cost, and sketch the marginal and average cost functions.
(3) A firm has cost function C(y) = 2y + 7. Sketch the marginal and average cost functions.
2. Introduction to Optimisation
In economic applications we often want to find the (global) maximum or minimum of a function. For example, we may want to find:
• the level of output where a firm’s profit is maximised
• the level of output where a firm’s average cost of production is minimised
• the number of hours per week that an individual should work, if he wants to maximise his utility.
Usually in economic applications the relevant function just has one stationary point, and this is the maximum or minimum that we want. But, having found a stationary point it is always wise to check that it is a maximum or minimum point (whichever is expected) and that it is a global optimum.
2.1. Profit Maximisation
Suppose that a firm producing y units of output has cost function C(y), and faces inverse demand function P (y). (In other words, if the firm makes y units, the total cost is C(y), and each unit can be sold at a price P (y).)
• The firm’s Revenue Function is: R(y) = yP (y)
• The firm’s Profit Function is: Π(y) = R(y) − C(y) = yP (y) − C(y)
• The firm’s optimal choice of output is the value of y that maximises its profit Π(y).
So we can find the optimal output by finding the maximum point of the profit function.
Examples 2.1: A monopolist has inverse demand function P (y) = 16 − 2y and cost function C(y) = 4y + 1.
(i) What is the optimal level of output?
The profit function is:
Π(y) = y(16 − 2y) − (4y + 1)
= 12y − 2y2− 1
Look for a maximum point where the derivative is zero:
dΠ
dy = 12 − 4y = 0 ⇒ y = 3
To check that this is a maximum look at the second derivative:
d2Π
dy2 = −4 so it is a maximum
If you think about the shape of the function (or sketch it), you can see that this the global maximum. So the optimal choice of output is y = 3.
(ii) How much profit does it make?
When y = 3:
Π(3) = 12 × 3 − 2 × 9 − 1 = 17 Profit is 17.
2.2. Marginal Cost, Marginal Revenue, and Profit Maximisation
We know that Marginal Cost is the cost of an additional unit of output, which is the derivative of the cost function:
M C(y) = dC dy
Similarly Marginal Revenue is the revenue obtained when one more unit of output is sold, and it is the derivative of the revenue function:
M R(y) = dR dy
To find the optimal level of output, we look for the point where dΠ dy = 0.
But we can write the profit function as:
Π(y) = R(y) − C(y)
⇒ dΠ
dy = dR
dy −dC dy So the optimal level of output is where:
dR
dy = dC
dy
⇒ Marginal Revenue = Marginal Cost
The optimal level of output for a profit-maximising firm is where dΠ
dy = 0, or (equivalently) where M R = M C
If you want to find the profit-maximising output you can either look for a stationary point of the profit function, or solve the equation M R = M C.
Exercises 6.2: Introduction to Optimisation (1)
(1) A competitive firm has cost function C(q) = 3q2+ 1. The market price is p = 24. Write down the firm’s profit as a function of output, Π(q). Find the profit-maximising level of output. How much profit does the firm make if it produces the optimal amount of output?
(2) A monopolist faces demand curve Qd(P ) = 10 − 0.5P . Find the inverse demand function Pd(Q), and hence the revenue function R(Q), and marginal revenue.
Suppose the monopoly wants to maximise its revenue. What level of output should it choose?
(3) A competitive firm produces output using labour as its only input. It has no fixed costs. The production function is y = 6L0.5; the market price is p = 4, and the wage (the cost of labour per unit) is w = 2. Write down the firm’s profit, as a function of the number of units of labour it employs, Π(L). How many units of labour should the firm employ if it wants to maximise its profit?
Further Reading and Exercises
• Jacques §4.6.
• Anthony & Biggs §8.1, 8.2.
3. More Rules for Differentiation 3.1. Product Rule
To differentiate the function:
y(x) = (5x + 3)(x2− x + 1)
we could multiply out the brackets and differentiate as before. Alternatively we could use:
The Product Rule: If y(x) = u(x)v(x) then y0(x) = u(x)v0(x) + u0(x)v(x)
or equivalently dy
dx = udv
dx + vdu dx
Examples 3.1: Product Rule
(i) Differentiate y(x) = (5x + 3)(x2− x + 1) Put u(x) = 5x + 3, and v(x) = x2− x + 1:
dy
dx = (5x + 3)(2x − 1) + 5(x2− x + 1)
= 15x2− 4x + 2
(ii) Differentiate 2x32(x3− 2)
f (x) = 2x32(x3− 2) Put u(x) = 2x32 and v(x) = x3− 2:
f0(x) = 2x32 × 3x2+ 3x12(x3− 2)
= 9x72 − 6x12 3.2. Quotient Rule
So far we have no method for differentiating a function such as:
y(x) = 3x − 2 2x + 1
but there is a rule for differentiating the quotient of two functions:
The Quotient Rule: If y(x) = u(x) v(x) then y0(x) = u0(x)v(x) − u(x)v0(x)
v(x)2 or equivalently: dy
dx = vdudx − udvdx v2
Examples 3.2: Quotient Rule (i) Differentiate g(x) = 3x − 2
2x + 1
Putting u = 3x − 2 and v = 2x + 1:
dg
dx = (2x + 1) × 3 − (3x − 2) × 2 (2x + 1)2
= 6x + 3 − 6x + 4
(2x + 1)2 = 7 (2x + 1)2 (ii) Differentiate h(x) = 3x
x2+ 2
h0(x) = (x2+ 2) × 3 − 3x × 2x
(x2+ 2)2 = 3x2+ 6 − 6x2
(x2+ 2)2 = 6 − 3x2 (x2+ 2)2 3.3. Chain Rule
To differentiate the function:
y(x) = 4(2x3− 1)7
we could multiply out the brackets to obtain a very long polynomial (of degree 21) and then differentiate. But this would take a very long time. Instead we can do it by writing y as a function of a function:
y = 4u7 where u = 2x3− 1 and then using the Chain Rule:
The Chain Rule
If y is a function of u, and u is a function of x:
dy dx = dy
du ×du dx Applying this to the example above:
dy
dx = 28u6× 6x2
= 28(2x3− 1)6× 6x2
= 168x2(2x3− 1)6
The Chain Rule is the rule for differentiating a composite function (see Chapter 4). Another way of writing it is:
Alternative Form of the Chain Rule If y(x) = v(u(x))
y0(x) = v0(u(x))u0(x)
Examples 3.3: Differentiate the following functions using the Chain Rule:
(i) y(x) = (1 − 8x)5
y = u5 where u = 1 − 8x dy
dx = 5u4× (−8) = 5(1 − 8x)4× (−8) = −40(1 − 8x)4 (ii) f (x) =p(2x − 3)
First write this as: f (x) = (2x − 3)12. Then:
f = u12 where u = 2x − 3 f0(x) = 1
2u−12 × 2
= 1
2(2x − 3)−12 × 2
= 1
√2x − 3 (iii) g(x) = 1
1 + x2
g = 1
u where u = 1 + x2 g0(x) = − 1
u2 × 2x
= − 2x
(1 + x2)2
(Note that we could have done this example using the quotient rule.) 3.4. Differentiating the Inverse Function
x y
............................. . .
. . .. . . . . .. . . . .. . . . . .. . . . . .. . . .
.. . . .. . . .. . . . .. . . .. . . . .. . . .. .
.. . .. . . .. . . .. . .. . . .. . .. . . .. . .
. .. . .. . .. . .. . .. . .. . .. . .. . . .. .
.. . .. . .. .. . .. . .. .. . .. . .. .. . ..
. .. .. . .. .. . .. .. . .. . .. .. . .. .. . ..
....
....
..
....
....
..
....
....
..
....
......
......
......
......
......
......
...
...
...
...
...
...
...
∆x
∆y
Remember from Chapter 5 that the derivative of a function is approximately equal to the ratio of the change in y to the change in x as you move along the function:
dy dx ≈ ∆y
∆x If a function has an inverse:
y = f (x) and x = f−1(y)
then the gradients of these two functions are:
dy dx ≈ ∆y
∆x and dx dy ≈ ∆x
∆y
So the gradient of the inverse function f−1 = 1
the gradient of the function f dx
dy = 1
dy dx
We will use this rule in later sections of this chapter.
Exercises 6.3: More Rules for Differentiation (1)
(1) Use the product rule to differentiate:
(a) y(x) = (x2+ 3x + 2)(x − 1) (b) f (z) = z(1 −√ z) (2) Use the quotient rule to differentiate: k(x) = x + 1
2x3+ 1
(3) Use the chain rule to differentiate: (a) a(x) = 2(2x − 7)9 (b) b(y) =p y2+ 5 (4) For a firm facing inverse demand function P (y) = 1
2y + 3, find the revenue function R(y), and the marginal revenue.
3.5. More Complicated Examples
Examples 3.4: Sometimes we need more than one of the product, quotient and chain rules:
(i) Differentiate the function y(x) = x3(x2+ 1)4. By the Product Rule: dy
dx = x3× d
dx(x2+ 1)4
+ 3x2× (x2+ 1)4 Then use the Chain Rule to differentiate the term in curly brackets:
dy
dx = x3× 4(x2+ 1)3× 2x + 3x2(x2+ 1)4
= 8x4(x2+ 1)3 + 3x2(x2+ 1)4 Factorising: dy
dx = x2(x2+ 1)3(11x2+ 3) (ii) Differentiate: f (x) =
x
x + α
β
First use the Chain Rule, with u = x+αx : f0(x) = β
x
x + α
β−1 d dx
x
x + α
Then use the Quotient Rule to differentiate the term in curly brackets:
f0(x) = β
x
x + α
β−1 α (x + α)2
= αβ xβ−1 (x + α)β+1
Exercises 6.4:
(1)
(1) Find the derivative of y(x) =
r x
1 − x (2) Differentiate z(t) = ta
(2t + 1)b where a and b are parameters.
(3) If f (q) = q5(1 − q)3, for what values of q is f (q) increasing?
(4) (Harder ) Using the chain rule and the product rule, derive the quotient rule.
Further Reading and Exercises
• Jacques §4.4.
• Anthony & Biggs §6.2, §6.4 and §6.5.
4. Economic Applications 4.1. Minimum Average Cost
If a firm has total cost function C(q), where q is its output, its average cost function is:
AC(q) = C(q) q
The derivative of the average cost function can be found using the Quotient Rule:
dAC
dq = qC0(q) − C(q) q2 At a point where average cost is minimised, dAC
dq = 0, so:
qC0(q) − C(q) = 0 C0(q) = C(q)
q
M C = AC
We have proved that for any cost function, M C = AC when AC is minimised: the marginal cost function passes through the minimum point of the average cost function.
4.2. The Marginal Revenue Product of Labour
If a firm uses labour to produces its output y, with production function f (L), revenue is a function of output, and output is a function of labour:
Revenue = R(y) where y = f (L)
The marginal revenue product of labour is the additional revenue generated by employing one more unit of labour:
M RP L = dR dL Using the chain rule:
dR
dL = dR
dy × dy dL
= M R × M P L
The marginal revenue product of labour is the marginal revenue multiplied by the marginal product of labour.
Exercises 6.5:
(1)
(1) A firm has cost function C(q) = (q + 2)β where β is a parameter greater than one.
Find the level of output at which average cost is minimised.
(2) A firm produces output Q using labour L. Its production function is Q = 3L13. If it produces Q units of output, the market price will be P (Q) = 54
Q + 9. (a) What is the firm’s revenue function R(Q)?
(b) If the firm employs 27 units of labour, find: (i) output (ii) the marginal prod-uct of labour (iii) the market price (iv) marginal revenue (v) the marginal revenue product of labour
5. Economic Application: Elasticity 5.1. The Price Elasticity of Demand
The gradient dqdp of a demand curve q(p) gives a measure of how “responsive” demand is to price: how much demand falls (it is usually negative) when price rises a little:
dq dp ≈ ∆q
∆p
But the gradient depends on the units in which price and quantity are measured. So, for example, the demand for cars in the UK would appear by this measure to be more responsive to price than in the US, simply because of the difference in currencies.
A better measure, which avoids the problem of units, is the elasticity, which measures the proportional change in quantity in response to a small proportional change in price. The price elasticity of demand is the answer to the question “By want percentage would demand change, if the price rose by one percent?”
Price elasticity of demand, = p q
dq dp
You can see where this definition comes from if we write:
p q
dq dp ≈ p
q
∆q
∆p = ∆q/q
∆p/p = proportional change in q proportional change in p
Examples 5.1: For the demand function q(p) = 10 − 2p:
(i) Find the price elasticity of demand.
dq
dp = −2
⇒ = p
q × (−2)
Substituting for q we obtain the elasticity as a function of price only:
= −2p 10 − 2p
(ii) Evaluate the elasticity when the price is 1, and when the price is 4.
Substituting into the expression for the elasticity in (i):
when p = 1, = −1
4 and when p = 4, = −4
In this example, demand is inelastic (unresponsive to price changes) when the price is low, but elastic when price is high.
We say that demand is:
• inelastic if || < 1
• unit-elastic if || = 1
• elastic if || > 1
Warning: The price elasticity of demand is negative, but economists are often careless about the minus sign, saying, for example “The elasticity is greater than one” when we mean that the absolute value of the elasticity, ||, is greater than one. Also, some books define the elasticity as −pqdqdp, which is positive.
5.2. Manipulating Elasticities
In the example above we started from the demand function q(p) = 10 − 2p, and obtained the elasticity as a function of price. But if we can find the inverse demand function, we could express the elasticity as a function of quantity instead:
q = 10 − 2p ⇒ p = 5 − q 2 Then:
⇒ = p
q × (−2) (as before)
= q − 10 q
We can also find the price elasticity of demand if we happen to know the inverse demand function p(q), rather than the demand function. We can differentiate to find its gradient dpdq. Then the gradient of the demand function is:
dq dp = 1
dp dq
(from section 3.4) and we can use this to evaluate the elasticity.
Examples 5.2: If the inverse demand function is p = q2− 20q + 100 (for 0 ≤ q ≤ 10), find the elasticity of demand when q = 6.
Differentiatiating the inverse demand function and using the result above:
dp
dq = 2q − 20 ⇒ dq
dp = 1 2q − 20
⇒ = p q
1 2q − 20
Here, it is natural to express the elasticity as a function of quantity:
= q2− 20q + 100 q(2q − 20) Evaluating it when q = 6:
= 16
6 × (12 − 20) = −1 3 5.3. Elasticity and Revenue
If a firm faces inverse demand function p(q), its revenue function is:
R(q) = p(q)q
As quantity increases, more items are sold, but at a lower price. What happens to revenue?
Differentiating, using the Product Rule, we obtain Marginal Revenue:
dR
dq = p + qdp dq
= p
1 +q
p dp dq
= p 1 + 1
p q
dq dp
!
= p
1 +1
= p
1 − 1
||
From this expression we can see that revenue is increasing with quantity if and only if || > 1:
that is, if and only if demand is elastic.
5.4. Other Elasticities
We can define other elasticities in just the same way as the price elasticity of demand. For example:
• If qs(p) is a supply function, the elasticity of supply is p qs
dqs dp
• If Y (L) is a production function, L Y
dY
dL is the elasticity of output with respect to labour.
Exercises 6.6: Elasticity (1)
(1) Find the price elasticity of demand for the demand function Qd(P ) = 100 − 5P , as a function of the price. What is the elasticity when P = 4?
(2) If the inverse demand function is pd(q) = 10+20q , find the price elasticity of demand when q = 4.
(3) Find the elasticity of supply, as a function of price, when the supply function is:
qs(p) =√ 3p + 4.
(4) Show that demand function qd(p) = αp−β, where α and β are positive parameters, has constant elasticity of demand.
Further Reading and Exercises
• Jacques §4.5.
• Anthony & Biggs §9.1.
• Varian “Intermediate Microeconomics” Chapter 15 discusses elasticity fully, without using calculus in the main text, and using calculus in the Appendix.
6. Differentiation of Exponential and Logarithmic Functions
6.1. The Derivative of the Exponential Function
The reason why the exponential function ex is so important in mathematics and economics is that it has a surprising property:1 the derivative of ex is ex.
If y = ex then dy dx = ex Using the chain rule we can show easily that if a is a constant:
If y = eax then dy
dx = aeax
Examples 6.1: Differentiate:
(i) 2e3x
y = 2e3x ⇒ dy
dx = 2 × 3e3x= 6e3x (ii) Y (t) = ae−bt
Y0(t) = a × (−b) × e−bt= −abe−bt (iii) k(x) = 5ex4+2x
Using the chain rule: ⇒ k0(x) = 5ex4+2x× (4x3+ 2)
= 5(4x3+ 2)ex4+2x 6.2. The Derivative of the Logarithmic Function
The logarithmic function ln x is the inverse of the exponential function, and we can use this to find its derivative:
y = ln x ⇒ x = ey dx
dy = ey = x dy
dx = 1
dx dy
= 1 x So we have obtained a rule for differentiating ln x:
If y = ln x then dy dx = 1
x
1You could try to verify this by differentiating ex= limn→∞ 1 +nxn
.
Examples 6.2: Differentiate the following functions:
Exercises 6.7: Differentiation of Exponential and Logarithmic Functions (1)
(1) Differentiate the function: f (x) = 12e8x. (2) What is the derivative of 4 ln(3x + 1)?
(3) Find the derivative of C(q) = 3qeq2+ 4e−2q.
(4) Differentiate the function: g(y) = y2ln(2y)+2. For what values of y is g increasing?
(5) Find and classify the stationary points of the function: F (t) = e−t(t2+ 2t − 2).
(6) If f (x) is any function, find the derivative of (a) y(x) = ef (x) (b) z(x) = ln f (x).
6.3. Economic Application: Growth
In Chapter 4, we modelled the GDP of a country as depending on time, according to the equation:
Y (t) = Y0e0.03t
(Remember that the parameter Y0 represents GDP at time t = 0.) Differentiating this equation we get:
Y0(t) = 0.03Y0e0.03t
= 0.03Y (t)
So we can now see that the rate of increase of GDP, Y0(t), is a constant proportion, 0.03, of GDP. In other words, GDP is growing at a rate of 3%.
6.4. Economic Application: The Optimum Time to Sell an Asset
F (t)
Suppose that, as a forest grows over time, the value of the timber is given by F (t), where F is an increasing, concave, func-tion of time. Then, if the interest rate is r, the present value of the forest at time t is:
V (t) = F (t)e−rt (See Chapter 3.)
The best time to cut down the forest and sell the timber is when the present value is max-imised. Differentiating (using the product rule):
dV
dt = F0(t)e−rt− rF (t)e−rt
= (F0(t) − rF (t))e−rt The optimum moment is when dV
dt = 0:
F0(t) = rF (t) ⇒ F0(t) F (t) = r
This means that the optimum time to sell is when the proportional growth rate of the forest is equal to the interest rate. (You can verify, by differentiating again, that this is a maximum point.)
To see why, call the optimum moment t∗. After t∗, the value of the forest is growing more slowly than the interest rate, so it would be better to sell it and put the money in the bank.
Before t∗, the forest grows fast: it offers a better rate of return than the bank.
Exercises 6.8:
(1)
(1) A painting is bought as an investment for £3000. The artist is increasing in popularity, so the market price is rising by £300 each year: P (t) = 3000 + 300t.
If the interest rate is 4%, when should the painting be sold?
(2) The output Y of a country is growing at a proportional rate g and the population N is growing at a proportional rate n:
Y (t) = Y0egt and N (t) = N0ent Output per person is y(t) = Y (t)
N (t). Using the quotient rule, show that the
N (t). Using the quotient rule, show that the