Nth-derivative test - practice materials PDF

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Procedure here is to undertake successive differentiation until a dn y derivative is obtained, i.e. n 0 at the value of x corresponding to stationary dx point. If: dn y dx n

0 and n is an EVEN number then y is a local

dn y dx n

0 and n is an EVEN number then y is a local

dn y dx n

0 and n is an ODD number then y is a

y

⇒

dy dx

3x 3

2x 2

9x2

dy 0 dx

5x

4x

5

⇒ 9x2 ⇒ x or x

2

4x 5 0 5 9 1

d2 y dx 2

18x

d2 y dx2

x

4

5

5 18  4 14 > 0 9

⇒

18 1

⇒

9

2

dy dx2

x

4

14 < 0

1

d2 y Are there any values of x such that 2 dx d2 y dx2 ⇒

0 ⇒ 18x

4

0 ⇒x

2 9

0?

: A straight line joining any two points on the function lies everywhere function except at the points themselves.

the

: A straight line joining any two points on the function lies everywhere function except at the points themselves. d2 y The condition for a (strictly) concave function is thus that dx 2 d2 y that for a (strictly) concave function is that 2 0, x dx

the

0, x, while

Note use of term ‘strictly’ with the above – where straight line joining points (a section of) the function, then function is described simply as concave/convex as appropriate, and inequalities above are replaced by and respectively

y

⇒

dy dx

3x 3

2x 2

9x2

4x

5x

2

5

d2 y ⇒ dx2

18x

d2 y For 2 dx

0 ⇒ 18x + 4 > 0 ⇒ 18x> 4

4

⇒x

2 9

d2 y For 2 dx

0 ⇒ 18x

4

0 ⇒ 18x< 4 ⇒ x

So function is convex for x

2 ; concave for x 9

2 9 2 9

2 is our (non-stationary) point of inflexion, which thus separates 9 concave and convex portions of function. Note that x

By definition:

TR(Q) TC(Q), which is maximised when the first order d dTR dTC 0 is satisfied, i.e. when MR = MC. condition given by dQ dQ dQ d2 d2TR d2 TC Second order condition for a maximum is: dQ2 dQ 2 dQ 2 dMR dMC (i.e. MC cuts MR ‘from below’). equivalent to: dQ dQ

0 , which is

If the opposite applies, then extremum is a of profits (see the diagram below which uses a cubic total cost function and a quadratic total revenue function)

TR, TC

Q

Q

MR, MC

Q

A monopolist has total cost function:

TC

100

2Q

1 2 Q 10

The inverse demand function is:

P

20- 1 Q 5

(a) Find the output (and hence price) at which the firm’s profits are maximised, checking that the second order condition is satisfied. 1 2 TR = PQ = 20Q – 5 Q

= TR – TC

= 20Q –

1 2 Q – (100 + 2Q + 5

= 18Q –

3 2 Q – 100 10

1 2 Q) 10

d 6 = 18 – Q=0 dQ 10

⇒ Q = 30, P = 20 – 1 (30) = 14 5

d2 dQ 2

6 10

0 ⇒ maximum

(b) For the above cost and demand functions, find the levels of output at which total revenue is maximised. Calculate the elasticity of demand at (i) the level of output which maximises profit and (ii) maximises total revenue. Comment.

TR = 20Q –

1 2 Q 5

dTR 2 = 20 – Q = 0 dQ 5 d2TR 2 = – dQ2 5

Ed =

(i)

0 ⇒ maximum

P dQ P 1 . . Q dP Q dP dQ max

⇒ Q = 50 (P = 10)

1 P . 1 Q 5

5

P Q

 14  : Ed = –5   = –2 1 , i.e. elastic 3  30 

 10  (ii) TRmax: E d = –5   = –1, i.e. unit elastic  50 

If TC = f(Q), then AC

f (Q) . The first order condition (by quotient rule) is given Q

by: dAC dQ

Qf ' (Q) f (Q) Q2

which

0

rearranges

to

give

f ' ( Q)

f ( Q) , Q

i.e. MC = AC. [Second order condition is more complicated. Second derivative is (again by quotient rule): d 2AC dQ

2

Q 2[ f ' ' (Q).Q

f ' (Q).1 f ' (Q)] 2Q[Qf ' (Q) Q4

f (Q)]

f ' ' (Q).Q3

2Q 2 f ' (Q) 2Qf (Q) Q4

f ' ' ( Q) Q

2f ' ( Q) Q2

2f ( Q) Q3

f ( Q) , so that Q f ( Q)and thus last term in second derivative can be rewritten to give:

However, from first order condition, we know that f ' (Q) Qf '(Q)

d 2AC dQ 2

f ' ' ( Q) Q

2f ' ( Q) Q2

2f ' ( Q) Q2

f ' ' ( Q) Q

For a minimum, second derivative must be positive, which will be true for all Q>0 provided f ' '( Q) 0 (i.e. total cost function is convex).]...