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# Problems on profit maximization and minimization of cost function

Applications of Differentiation maxima and minima

Problems on profit maximization and minimization of cost function:

Example 6.27

For a particular process, the cost function is given by C = 56 - 8x + x2 , where C is cost per unit and x, the number of unitâ€™s produced. Find the minimum value of the cost and the corresponding number of units to be produced.

Solution :

C = 56 - 8x + x2

Differentiate with respect to x, The minimum value of cost = 56â€“32+16

= 40

The corresponding number of units produced = 4

Example 6.28

The total cost function of a firm is C(x) = x3/3 â€“ 5x2 + 28x + 10 where x is the output. A tax at the rate of â‚ą 2 per unit of output is imposed and the producer adds it to his cost. If the market demand function is given by p = 2530 â€“ 5x, where p is the price per unit of output, find the profit maximizing the output and price.

Solution :

Total revenue:  R = p x

= (2530 â€“ 5x)x

= 2530xâ€“5x2

Tax at the rate â‚ą 2 per x unit = 2x. = 2530 â€“ 5(50)

= â‚ą 2280.

Example 6.29

The manufacturing cost of an item consists of â‚ą 1,600 as over head material cost â‚ą 30 per item and the labour cost â‚ą a (x2 /100) for x items produced. Find how many items be produced to have the minimum average cost.

Solution :

As per given information for producing x units of certain item C(x) = labour cost + material cost + overhead cost AC is minimum at x = 400

Hence 400 items should be produced for minimum average cost. Tags : Applications of Differentiation maxima and minima , 11th Business Mathematics and Statistics(EMS) : Chapter 6 : Applications of Differentiation
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11th Business Mathematics and Statistics(EMS) : Chapter 6 : Applications of Differentiation : Problems on profit maximization and minimization of cost function | Applications of Differentiation maxima and minima