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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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
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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
dq
dp
2)31(
15)155(
q
2)31(5q
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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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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
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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