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Exercises - Math Problem Example

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Math Exercises
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Exercises

Download file to see previous pages... 1 A firm manufactures and sells q units of a product at price = £(575 – ½ q)
which has unit costs of £(q2 – 25q) and fixed costs of £45,000.

(a) Write down expressions for: revenue, profit and average cost
in terms of output(q) of the firm. [1 mark]

Revenue = (575 – ½ q ) q
= 575q – ½ q2

Profit = Revenue – Total Cost

= 575q – ½ q2 - [(q2 – 25q )q +45,000]

= 575q – ½ q2 – q3 + 25q2 - 45,000

= – q3 + 24.5q2 + 575q - 45,000

Average Cost = Total Cost / q

= q2 – 25q + 45,000/q


(b) Find expressions for: marginal revenue, marginal cost, marginal
profit and marginal average costs in terms of output (q). [2 marks]


Marginal Revenue, Marginal Cost, Marginal Profit and Marginal Average Costs is the derivative of Revenue, Cost, Profit, Average Costs . Since the derivative of f(x) = xn is nxn-1, we have:

Marginal Revenue = 575 – q

Marginal Cost = 3q2 -50q

Marginal Profit = -3q2 + 49q +575

Marginal Average Cost = 2q – 25 -45,000/q2 (since 1/q = q-1)

(c) Find the output levels of the firm that and confirm that the output levels found do indeed maximise or minimise these functions [ 1 mark]


(i) Maximise revenue

















• This is the graph of Revenue = 575q – ½ q2 , we can see that it is maximised at q = 575.


(ii) Minimise costs

To minimise costs, set marginal costs to 0

q = 50 / 3 or approx 17 units

















This is the graph of Costs = q3 - 25q2 + 45,000. We can see that the minimise value is approximately at q =17.


(iii) Maximise profits

To maximise profits, set marginal profits to 0

-3q2 + 49q +575 = 0

Using the quadratic formula, we have:

q = 23.23 , -7.89

Disregarding the negative value, we have:

q = 23 units.















This is the graph of Profit = -q3 + 24.5q2 + 575q - 45,000. We can see that the maximum value is approximately at q=23.


(iv) Minimise average costs

To minimise average costs, set marginal average costs to 0:


2q - 25 -45,000/q2 = 0 (multiply both sides by q2)

2q3 - 25q2 - 45,000 = 0

With the use of trial and error, we get the only possible value as:

q = 33 units.







This is the graph of Average Cost = q2 - 25q + 45,000/q. We can see that the maximum value is approximately at q=33.


2. The demand function for a product is given by the following expression:

q = 25 + 200
(p - 2)

(a) Calculate the demand at prices 3 and 7 [1/2 mark ]

For p = 3:

q = 25 + 200
(3 - 2)
q = 25 + 200

q = 225

For p = 7:

q = 25 + 200
(7 - 2)
q = 25 + 40

q = 65


Answer in (Q,P) form: (225,3), (65,7)


(b) Calculate the ARC elasticity of demand with respect to price
between the prices given in part (a) and comment on whether
demand is elastic or inelastic between these prices. [1/2 mark]

Earc = (Q2-Q1) / [(Q2+Q1)/2]
(P2-P1) / [(P2+P1)/2]

Earc = (65-225) / [(65+225)/2]
(7-3) / [(7+3)/2]

Earc = -160 / 145
4 / 5

Earc = -40 = -1.38
29

Since an "elastic" good is where price elasticity of demand is greater than one, we can consider that the demand is elastic between these prices.

(c) Find an expression for POINT elasticity of demand with respect
to price in terms of price. [ 1 mark]

Ept = (q/ p) * p/q

The derivative of q = 25 +200/(p-2) is

q/ p = 0 + -1 (200) (p-2)-2

And q = 25 +200/(p-2)

Hence:

Ept = [-200p/ (p-2)2]/ [25 +200/(p-2)]



(d) Calculate POINT elasticity of demand at prices 3 and 7 and
comment on their values and on the relationship between
ARC and POINT elasticity [1/2 mark]
Ept = [-200p/ (p-2)2]/ [25 +200/(p-2)]

Ept (3) = (-600/ 1)/ 225

= -2.67

Ept (7) = -56/ 65

= -0.862

The value of arc elasticity is in between the value of point elasticity which is expected ...Download file to see next pagesRead More
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