buss math 261112

7/27/2019 Buss Math 261112

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The demand function : P = 17 – X the supply function : P = ¼ X + ¾

Find the equilibrium and how much is the subsidy must given by the

government so as the comodity is free from charge (gratis)

Solution

Demand : X = 17 – P , Supply : X = 4P – 3 P = 4 , X = 13

Equilibrium : (13 , 4)

Suppose subsidy given is x to get free

Supply : P = ¼ X + ¾

Because of the subsidy, the function becomes P = ¼ X + ¾ - x

X = 4P – 3 + 4x theThe equilibrium after subsidy : X’S = X’D

4P – 3 + 4x = 17 – P. Free means P = 0 4x = 20 x = 5

So the government must give subsidy about 5 to make free price

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TWO COMMODITIES

Two comodities that is substitutive or complimentary

the procedure is the same with single commodity

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DIFFERENTIAL

Differentiation : Concerned with measuring the effect on the value of afunction when there is an infinitely small change in the value of a

variable in the function

Eg. How much does D or S change as prices drop?

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What we measured

The slope of a line at a particular point on the line

This is also known as the first derivative: Effectively measuring the slope or gradient of Y at a

particular value X i If y = f( x) then

Also shown as f ‘ (x)

x

y

dx

dy

x

0lim

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5

Rules of Differentiation

1. Constant function f(x) = k => f ' (x) = 0

Fixed Costs: f(q) = 100

What does this function look like?

=> f ' (q) = 0

Ie. Fixed costs do not change with small changes inoutput

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2. Linear fn f(x) = a + bx => f ' (x) = b f(Q) : P = 20 – 2 Q

Sketch this function

What can this function represent in Economics?

Find f ' (Q) = -2

What does this represent?

= slope of demand line,

For each increase in Q by 1, price needs to fall by $2

Or for each increase in P by $2, quantity falls by 1 unit Assumes all other factors affecting demand are held constant

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3. Power fn f(x) =kxn

=> f ' (x) = nkxn-1

eg C = 5Q2

=> f ' (Q) = 2*5*Q2-1 =10Q

What if C = 4Q0.7 what is f ' (Q) ? f ' (Q) = 4*0.7*Q0.7-1 = 2.8Q-0.3

What is f ' (Q) = dC/dQ known as in Economics?

= marginal cost = MC

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Sometimes easier to find dx/dy instead of dy/dx, so can just invert:

)(

1

dydxdx

dy

7/27/2019 Buss Math 261112

http://slidepdf.com/reader/full/buss-math-261112 9/25Econ204 topic 1 9

4. Sums and Differences

f (x) = g(x) ± h(x)

=> f '(x) = g '(x) ± h ' (x)

Eg TC = Q2 + 5Q +20

Then dTC/dQ = MC = 2Q + 5

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Relationship between AR (Average Revenue), MR (Marginal

Revenue) & TR (Total revenue)

Demand : P = 20 – 2 Q =ARSlope of AR = -2

What is the Total Revenue (TR) obtained for any unit sold?

TR = P.Q

Hence TR = (20 –2Q)*Q

=> TR= 20Q – 2 Q2

What is the marginal Revenue (MR) for each extra unit sold?

MR = dTR/dQ =f’(Q)

MR =20 – 4 Q

=> slope = - 4, twice slope of AR

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5. Products :

let f(x) = g(x)*h(x) where g(x) and h(x) are both

differentiable f ' (x) = g(x)*h ' (x) + h(x)*g ' (x)

Or let u = g(x) and v = h(x)

Then f ' (x) = u *dv/dx + v*du/dx

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Example

Assume same demand fctn : P = 20 –2Q

Find d(PQ)/dQ using product rule

Product rule =>

Note: need P and Q functions to both be in terms of Q

dQdP Q

dQdQ P

dQTRd )(

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Answer

Substitute in values into above egn.

=(20-2Q)*(1) + (Q)*(-2)

= 20 – 2Q -2Q=20 – 4Q

This is the same as if you had used TR = P*Q and substituted for P egn written in terms of Q

dQ

dP

QdQ

dQ

P dQ

TRd

)(

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Alternative method

Could find Marginal Revenue (MR) by just differentiating

TR = P*Q

= (20 -2Q)*Q

= 20Q – 2Q2

Hence using earlier rules

QdQdTR MR 420

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6. Let f(x) = g(x)/h(x) where both are differentiable

f ' (x) = [h(x)*g ' (x) - g(x)*h ' (x)] / [h(x)]2

Eg. If

dq

dp f ind

q

q p

31

5

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Let g(q) = 5q and h(q) = 1-3q

Then

2)31(

)3(55)31(

q

qq

dq

dp

2)31(

15)155(

q

qq

2)31(5q

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17

7. fn f(x) = [g(x)]n where g(x) is differentiable and n isany real number

Then can’t use just the power rule as g(x) is diff

Then f ' (x) = n[g(x)]n-1 *g ' (x)Eg.

2)53( QC

3.)53(2 12 QdQdC )53(6 Q 3018 Q

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8. used for composite functionsSuppose y = f(u) and u = g(x)

=> y = f[g(x)]

Then dy/dx = dy/du * du/dx

Eg 2)53( QC 53 Qu

dQ

du

du

dC

dQ

dC .

30183).53(2 QQ

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Another example

If y = w3/4 and w = x 2 – x +1

find dy/dx

note:

dx

dw

dw

dy

dx

dy.

4

3 4/1

w

dw

dythen

12 xdx

dwand

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20

12.

4

3 4/1

xw

dx

dy so

1214

3 25.02

x x xor

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9. Logarithms

To find a log of a number

Number = base power

Then: log base (number) = power

E.g. X = e y then loge x = y = ln x

Y = ln X dY/dX = 1/X

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10. For an exponential function eg : Y = e X

Then dY/dX = e X

Same rules for compound fns eg Y = e3X

Let u = 3X => dY/dX = eu *3 = 3e3X

Present values with interest compounding continuously

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Optimisation

How do we know when a function has been maximised or minimised?

Depending on shape of function there could be more than 1 ‘hill’ or‘valley’, or extrema

Optimal result = most extreme position, relative to other extremaand end pts of fctn.

We will only cover unconstrained optimisation in this paper, notconstrained optimisation.

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Mathematically, at a turning point, (ie max or min), the gradient(slope) of the function will be 0

Necessary condition for a turning point is that

dy/dx = f ' (x) = 0

To determine whether max or min need to consider second orderderivative

7/27/2019 Buss Math 261112


Problems

The profit of radio selling

P(x) = 400(15-x)(x-2) where x is price of each. What is the maximum price from the company

Solution

P(x) = -400 x2 +6800 x – 12000

P’(x) = - 800 x +6800

Maximum means P’(x) = 0 0 = -800x + 6800 x = 8.5

The maximum price is 8.50 each

buss math 261112

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