lecture 0 math for managerial economics

8/12/2019 Lecture 0 Math for Managerial Economics

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Outline:

1. Functions.2. Tangents and slopes.3. Derivatives: Meaning and computation.4. Derivatives with multivariable functions.

1. Functions:

Single variable: A single variable function gives the level ofone variable (y) as determined by another variable (x). The

defining feature of a function is that a value of x implies aunique value of y.

Notation: Two standard notations: (i) y = f(x). (ii) f = f(x).

In notation (i), the function value is given another label (y),beside f. In notation (ii), the function value is just called onething (f).

Examples:

General form examples: Suppose that a firm’s profits ( π ) aredetermined by the quantity the firm sells (q). Then we havethe functional relationship: π = π (q). (I’m using notation (ii)that economists usually use.)

Suppose that a firm’s total costs (C) are determined by thequantity (q) the firm produces. Then we have the functionalrelationship C = C(q), again using notation (ii).



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Specific form examples: 2π 1000q q= −= −= −= −

C = 500 + 4q +q 2

C = 300 + 2q (straight line example)

Graphing Functions: The determining value is usuallymeasured on the horizontal axis, and the functional value isusually measured on the vertical axis. Other terminology forthese two values is, respectively, the independent variable(determining value) and the dependent variable (functionalvalue).

We’ll focus on continuous (no jumps) and smooth (no kinks)functions.

Examples are graphed on next page.



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Figure 0.1: Functions Graphed

ΠΠΠΠ

C

C

q

q

q



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Multivariable Functions: A multivariable function gives thelevel of one variable (y) as determined by more than one othervariable (x 1,x2,…,xn). The defining feature of such a function isthat a set of values of the x i’s implies a unique value of y. Thexi’s are frequently referred to as the independent variables andthe function value (y) as the dependent variable.

Notation: Again, two standards: (i) y = f(x 1,x2,…,xn)(ii) f = f(x 1,x2,…,xn)

General Form Examples: Total profits ( π ) in a market withtwo sellers might depend on the level of each of their outputs,

q1 and q 2. Thus: π = π (q1,q2).

The price (P) at which a product can be sold might depend onthe total quantity that is sold (q) and the average level of buyerincomes (I). Then: P = P(q,I).

The quantity of a firm’s output (q) might depend on the levelof labor employed (L) and the level of machine inputsemployed (K). Then: q = q(L,K).

Specific Form Examples:

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2 22 1 2 1 2 1 2

π (q ,q ) 1000(q q ) q q 2q q= + − − −= + − − −= + − − −= + − − −

P(q,I) = 200I – 2q 2



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Graphing Multivariable Functions:

We won’t much be concerned with this, except for onepossibility discussed in a moment.

But note these things: To graph a function, you need an axisfor the functional value and each independent variable. Ifthere are more than two independent variables, then it isimpossible to graph the entire function since we only havethree dimensions of space. If there are two independentvariables, then, if you’re talented, you can draw a threedimensional picture. We won’t need to do this.

Sometimes we’ll have a multivariable function and we’ll beinterested in just the relationship between one of theindependent variables and the function value. We can find thisrelationship if we fix the other independent variables. Ineffect, then, we have a single variable function that we cangraph.

Example: Suppose that price (P) depends on the quantity sold(q) and average consumer income as in the example above:

P(q,I) = 200I – 2q 2

If we fix I at say I = 50,000, then we have a functionalrelationship between q and P: P = (200)(50,000) – 2q 2.We could obviously graph the latter.

We could as well find the functional relationship between qand P for a different value of I, say I = 75,000. This would begiven by: P = (200)(75,000) – 2q 2. Again, we could graph this.



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2. Tangents and Slopes

A straight line function has a slope. The general form of astraight line function is: y = a + bx, where a and b areconstants. An example would be: y = 100 + 5x.

The slope of a straight line function is the ratio of the change iny to a change in x. This ratio will be the same for any changein x. (This is generally written as ∆ y / ∆ x.)

For the general form y = a + bx, the slope will be given by b.(This is easy to prove.) For the example y = 100 +5x, the slope

would be 5.

The slope has a very useful interpretation. It tells you the rateat which the functional value (y) is changing as theindependent variable (x) is changed. For example, if the slopeis 5 and x rises by 1, then y would rise by 5 (i.e., the change in yequals the change in x times the slope). If x were to rise insteadby 2, then y would rise by 10.



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Now consider a non-linear function. Refer to the example inFigure 0.2 of a profit function. Again, we will be interested inhow the functional variable changes as the independentvariable changes. We’ll explain further why we’re interestedin this in a few minutes.

Consider this question: At what rate does profit change asquantity is changed? When the function is non-linear, theanswer depends on where you are on the function. At q = q 1,the rate at which profit changes as q changes is given by theslope of the “tangent line” at that point on the function. At q =q2, the rate at which profit changes is given by the slope of the

tangent line at that point on the function. The tangent line at apoint on a function is the straight line that just touches butdoesn’t cross the function.

The rate at which a function value changes as the independentvariable changes – the slope of the tangent line – is the“derivative” of the function.

Sometimes the derivative is referred to as the “instantaneousslope.”



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Figure 0.2: Tangent lines and rates of change

ΠΠΠΠ

q 1 q 2 q

Tangent line at q=q 2 ; itsslope equals the rate ΠΠΠΠ changes as q changeswhen q=q 2 .

Tangent line at q=q 1 ; its

slope equals the rateΠΠΠΠ

changes as q changeswhen q=q 1 .

.

.



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3. Derivatives: Meaning and Computation:

Suppose that I have a function, say profits as a function ofquantity produced, and I can compute its derivative. Whatgood does this do me??

There are two related applications:

Application 1 of Derivatives: Solving optimization problems.Refer to the upper panel of Figure 0.3, which graphs the profitfunction. The firm wants to find the output where profits areat a maximum. We can see in the graph that the output that

leads to profit maximization is q*. (We’ll typically use a “*” toindicate an optimal value.) How can we compute q* if we havethe function?

Note that the maximum has the property that the slope of the corresponding tangent line – i.e., the derivative – is equal to 0 atq* . This points to a method to find q*. If we compute the

derivative of the function and set it equal to 0, this will give us anequation in q, the solution of which will be q*. You’ll see this

precisely in a bit when we do an example.

Note, well, that this approach also works when our objective isto minimize a function. The lower panel of Figure 0.3 depictsan example of total cost (C) as a function of the ratio (R) of twotypes of labor inputs. The firm would want to hire the inputsin the ratio that minimizes total cost of producing. We can seein the graph that the cost minimizing ratio is R*. At R*, the

slope of the tangent line is 0 – the derivative is 0. If we can compute the derivative of the function, then we can set it equal to0 and solve for R*.



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Figure 0.3: Derivative Equals 0 at Maximum and Minimum

ΠΠΠΠ (q)

C(R)

q*

R* R

q

.

.

derivative of C(R)is 0 here

derivative of ΠΠΠΠ (q)

is 0 here



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I should mention that there are some pitfalls associated withthe basic technique to solve optimization problems, but we’ll

just deal with them when we need to do so. (Briefly discuss anexample of this.) Almost without exception, we’ll deal inenvironments where the pitfalls are a non-issue.

Application 2 of Derivatives: Generally, the derivativeprovides useful information, this depending on the particularapplication. We’ll develop this kind of application throughoutthe course, but let me just mention a couple of examples tobegin to clarify the idea.

Take, again, the example of profits as a function of quantity.Suppose the firm is producing a particular quantity, say q 1.Suppose that the derivative of profits at q = q 1 is positive. Thismeans that the rate of change of profits as q increases ispositive. This means that the firm wants to produce more – itwould increase profits by doing so!

Consider total cost of a firm and suppose that it is a function ofquantity produced (which, of course, makes sense).Economists have a name for the derivative of this total costfunction: It’s called marginal cost. Marginal cost is the rate atwhich total cost changes as quantity changes. As we will see,marginal cost is a useful measure of cost. Why? If I can sellthe next unit produced for $100 and thus increase revenues by$100 – and if I know marginal cost equals $60 – then Icertainly want to produce the next unit and sell it. I wouldincrease my profits by the difference, $40. (We’ll discuss thisexample much more thoroughly later.)



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Two basic rules of differentiation:

Rule 1. Differentiating a One-Term Polynomial:

b b 1dyy ax , where a,b are cons tan ts bax

dx−−−−= → == → == → == → =

Two important special cases of the latter rule are:

(i) dyy ax or b 1 a

dx= = → == = → == = → == = → =

(ii) dyy a or b 0 0

dx= = → == = → == = → == = → =

Rule 2. Differentiating a Sum of Functions:

1 2 n1 2 n

df df df dyy f (x) f (x) ... f (x) ....

dx dx dx dx= + + + → = + + += + + + → = + + += + + + → = + + += + + + → = + + +

In words, Rule 2 is that the derivative of a sum offunctions is the sum of the derivatives of the individualfunctions.

With the latter two rules we can differentiate anypolynomial. Examples:

f(x) = 10 + 2x – 4x 2 2 1df 0 2 2 4 x 2 8x

dx−−−−→ = + − ⋅ ⋅ = −→ = + − ⋅ ⋅ = −→ = + − ⋅ ⋅ = −→ = + − ⋅ ⋅ = −



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f(x) = 12x + x 4

4 1 3df 12 4x 12 4x

dx−−−−→ = + = +→ = + = +→ = + = +→ = + = +

f(x) = 6xdf

6dx→ =→ =→ =→ =

(function is a straight line; rate of changeis a constant, i.e., the same everywhere on the function)

f(x) = 7df

0dx

→ =→ =→ =→ = (f is always 7, it doesn’t really depend on x,

so the rate of change of f as x changes is always 0)

Examples of Solving Optimization Problems:

Example 1: Recall the example of profits as a functionof the quantity produced:

π (q) = 1000q – q 2

At what q are profits maximized? Compute the

derivative and set it equal to 0, and then solve for q:setdπ

1000 2q 0.dq

= − == − == − == − =

The solution is q* = 500. The corresponding profits are:2π (500) 1000 500 500 250,000.= ⋅ − == ⋅ − == ⋅ − == ⋅ − = (You can’t get profits

any higher than 250,000!)



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Example 2: Suppose that total cost as a function of theoutput produced is:

C(q) = 500 + 4q + q2

It is a silly problem to think about minimizing total costwith respect to the output produced, because then youwould just produce nothing. But we might beinterested in finding the output level where the cost perunit of output or average cost is at a minimum. Averagecost (AC) is equal to total cost divided by q. Thus:

21500 4q q 500

AC(q) 4 q 500q 4 qq q

−−−−+ ++ ++ ++ += = + + = + += = + + = + += = + + = + += = + + = + +

This is a polynomial so we can find the derivative:

2 set2

dAC 500( 1)(500)q 0 1 1 0

dq q−−−−= − + + = − + == − + + = − + == − + + = − + == − + + = − + =

Solving the latter equation gives:

q * 500 10 5 22.36.= = ≈= = ≈= = ≈= = ≈

Computing average cost at this output gives:

AC(22.36) = 500·(22.36)-1

+ 4 + 22.36 ≈ 48.72

That’s as low as you can get average cost.



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The other type of application of using derivatives to “provideuseful information” we’ll see plenty of examples as the courseunfolds.

Three More Rules of Differentiation:

You won’t need these though they might at times allow aquicker derivation (obviously if you know them). I might usethem a very few times in getting some results in class. I’m notgoing to go through them, but they’re here is you want to learnthem.

Rule 3. Differentiating a Product of Functions:

(((( )))) 2 11 2 1 2

df df df f x f (x) f (x) f (x) f (x)

dx dx dx= ⋅ → = ⋅ + ⋅= ⋅ → = ⋅ + ⋅= ⋅ → = ⋅ + ⋅= ⋅ → = ⋅ + ⋅

Rule 4. Differentiating a Quotient of Functions:

1 22 1

12

2 2

df df f (x) f (x)

f (x) df dx dxf(x)f (x) dx f (x)

⋅ − ⋅⋅ − ⋅⋅ − ⋅⋅ − ⋅

= → == → == → == → =

Rule 5. Differentiating a Function of a Function of x (ChainRule):

1 21 2

2

df df df f (x) f (f (x))

dx df dx= → = ⋅= → = ⋅= → = ⋅= → = ⋅



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4. Derivatives with Multivariable Functions

When we have a multivariable function, we are sometimesinterested in how the function changes when just one of thevariables changes. Recall the example we mentioned where theprice (P) a firm can get depends on the quantity it sells (q) andaverage consumer income (I): P = P(q,I). We might beinterested in how price changes when just the quantitychanges. This only makes sense if we then fix the averageconsumer income. We can then calculate the derivative of Pwith respect to q in the usual way, where we treat I as aconstant. This is called a partial derivative, but it is just the

ordinary derivative where the other variable is held fixed.

Notation: When you have a multivariable function:f = f(x 1,x2,…,xn) and you take the derivative with respect to

one of the variables, say x 2 (thus holding the other x i’s

constant), it is written:2

f .

x

∂∂∂∂

∂∂∂∂

The interpretation is analogous to the ordinary derivative: Itis the rate of change of the function value as the x variable inquestion changes (holding other x variables fixed).



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Example Computation: Recall the example earlier of demandwhere the price (P) depends on the quantity sold (q) andaverage consumer income (I):

P(I,q) = 200I – 2q 2

Suppose that we want to find how price changes as quantitychanges, thus necessarily holding income (I) fixed. Computethe derivative of the equation with respect to q, treating I as aconstant value (which means treating I as if it were a number).We get:

P 0 4q 4qq

∂∂∂∂ = − = −= − = −= − = −= − = −∂∂∂∂

(the 0 because the first term, 200I, is

treated as a constant value and thus doesn’t vary with q)

Suppose we want to find how price varies with averageconsume income, then treating q as a constant. We get:

P200 0 200

I

∂∂∂∂= + == + == + == + =

∂∂∂∂ (the 0 by the same logic as in the previous

example).

Note that most of the problems we encounter will not bemultivariable, but some important problems are multivariable.



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Practice Problems (answers on the next page).

Compute the derivative:

1. y = 4 + 7x + 2x -2

2. 46y 17x

x= + (Hint: First rewrite as a polynomial. 1)

3. 22

(2 x)y 12 2x

x

+= + + (Hint: First rewrite as a polynomial. 2)

4. y = (4+x)(2+x) 2 (Hint: First rewrite as a polynomial. 3)

Find the maximum:

5. Let q denote quantity and P denote price. Because demand is downward sloping, theprice a firm can get declines with the quanitity; specifically: P = 100-2q. The firm’s totalcost is q 2.

(i) Profits equal revenues (P·q) minus total cost, but the price needs to be consistent withthe quantity sold through the demand curve. Exploiting this consistency requirement,write out profits as just a function of q. Let’s denote this function π (q).

(ii) Find the quantity that maximizes profits. Then calculate the corresponding price thatis being charged and the level of maximized profits.

6. Do the same problem but with demand P = 400-q and total cost 2q2.

1 If you know all the rules of differentiation, then you need not first rewrite as a polynomial.2 If you know all the rules of differentiation, then you need not first rewrite as a polynomial.3 If you know all the rules of differentiation, then you need not first rewrite as a polynomial.



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Answers to Practice Problems:

1. 3dy7 4x

dx−= −

2. Rewrite first as: 1 4y 6x 17x−

= +

Then: 2 3dy6x 68x

dx−= − +

3. Rewrite: 2 2 2 1 2y (2 x)x 12 2x 2x x 12 2x− − −= + + + = + + +

Then: 3 2dy4x x 4x

dx− −= − − +

4. Rewrite:2 2 2 3 2 3y (4 x)(2 x)(2 x) (4 x)(4 4x x ) 16 16x 4x 4x 4x x 16 20x 8x x= + + + = + + + = + + + + + = + + +

Then: 2dy20 16x 3x

dx= + +

5. (i) 2 2 2π (q) P(q) q q (100 2q)q q 100q 3q= ⋅ − = − − = −

(ii) Find d π /dq and set equal to 0. We have:

set *dπ 100 6q 0 Solving : q 50 / 3.dq

= − = =

Find P using the demand equation: P* = 100 – 2(50/3) = 200/3.

Compute profits at the maximum:2 2

π * P*q * (q*) (200 /3)(50 /3) (50 /3) ... 2500 /3.= − = − = =

6. (i) 2 2 2π (q) P(q) q 2q (400 q)q 2q 400q 3q= ⋅ − = − − = −

(ii) Find dπ

/dq and set equal to 0. We have:

setdπ

400 6q 0 Solving : q* 200 / 3dq

= − = =

Find P using the demand function: P* = 400 – 200/3 = 1000/3. Compute profits:



2 2π * P *q * 2(q*) (1000 / 3)(200 / 3) 2(200 / 3) ... 40, 000 / 3.= − = − = =

lecture 0 math for managerial economics

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