Question

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4. Suppose that a firm produces two different outputs, the quantities of which are repre- sented by yı and y2. In general, the firm's total costs can be represented by C(y1, y2). This function exhibits economies of scope if C(0,42) + C(4,0) > C(y1, y2) for all output levels of either good. 4.1 Explain in words why this mathematical formulation implies that costs will be lower in this multiproduct firm than in two single-product firms producing each good separately. 4.2 If the two outputs are actually the same good, we can define total output as y = y + y2. Suppose that in this case average cost decreases as y increases. Show that this firm also enjoys economies of scope under the definition provided here.

Answer 1

​​​​​​​​​​4.1. Given a firm produces two different products

• y1 is the quantity of first good
• y2 is the quantity of second good

Firms total cost is c(y1,y2) this shows that cost is the function of quantity of two goods produced

This mathematical formulation implies that cost will be lower in single multi product firm than in two different firms producing each goods separately. This is because when multiple goods are produced in single firm then there is optimal utilisation of its resources like machines and labours and secondly when two different firms will be set up for two goods then the total fixed cost (TFC) of firm will rise. Both of these factors lead to lower down the cost of two goods produced in a single firm.

4.2. Ac to the question a firm exhibits economies of scope if C(0,y2)+C(y1,0) > C(y1,y2) if two goods are actually the same and total output is y=y1+y2 if average cost decreases as output increases then this firm also enjoys economies of scope as when y increases then average cost decreases this is seen in a firm where two goods are produced in same firm and this is the situation required for economies of scope.

The conclusions of 4.1 can also be used to explain why this firm enjoys economies of scope as in single product firm cost decreases as output (y) increases and hence C(0,y2) + C(y1,0) is greater than C(y1,y2).