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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 - 8*x* + *x*^{2}
, 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 - 8*x* + *x*^{2}

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*) = x^{3}/3 â€“ 5x^{2}
+ 28*x* + 10 where x is the output. A
tax at the rate of â‚¹ 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€“ 5*x*, 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€“ 5*x*)*x*

= 2530*x*â€“5*x*^{2}

Tax at the rate â‚¹ 2 per *x* unit = 2*x*.

= 2530 â€“ 5(50)

= â‚¹ 2280.

**Example 6.29**

The manufacturing cost of an item
consists of â‚¹ 1,600 as
over head material cost â‚¹ 30 per
item and the labour cost â‚¹ a (*x*^{2} /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

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