PROBLEMS ON FIXED COST AND VARIABLE COST ON INTEGRATION

Problem 1 :

The fixed cost of a new product is $18000 and the variable cost per unit is $550. If the demand function p(x) = 4000 - 150x, find the break even values.

a)  15, 8    b)  7, 12    c)  3, 17    d)  5, 15

Solution :

Demand function is

p(x) = 4000 - 150x

Revenue function = demand function (number of units produced)

Revenue function = p(x) · x = (4000 - 150x) · x

R(x) = 4000x - 150x²

Total cost = fixed cost + variable cost

Number of units produced = x

= 18000 + 550x

Break even point will be there when revenue function and cost function will be equal.

18000 + 550x = 4000x - 150x²

150x² + 550x - 4000x + 18000 = 0

150x² - 3450x + 18000 = 0

Divide the equation by 150, we get

x² - 23x + 120 = 0

(x  -15) (x - 8) = 0

x = 15 and x = 8

The breakeven point is at 15 and 8. So, option a is correct.

Use the data from problem 2 to 4.

A company sells its product at $60 per unit. Fixed cost for the company is $18000 and the variable cost is estimated to be 25% of the total revenue.

Problem 2 :

Determine the total revenue function

a)  70x     b)  60x     c)  90x     d)  100x

Solution :

Let x be the number of units produced.

Selling price = 60

Revenue = Number of units produced · Cost per unit

Revenue = 60 x

So, option b is correct.

Problem 3 :

Determine the total cost function

a) 19000 + 6x    b)  20000 + 10x    c)  18000 + 15x

d)  4000 + 5x

Solution :

Total cost = Fixed cost + variable cost

Fixed cost = 18000

Variable cost =  25% of 60x

= 0.25(60x)

= 15x

Total cost = 18000 + 15x

So, option c is correct.

Problem 4 :

Determine the breakeven point 

a)  600      b)  400   c)  700    d)  1000

Solution :

Cost function = Revenue function

18000 + 15x = 60 x

18000 = 60x - 15x

18000 = 45x

x = 18000/45

x = 400

So, the breakeven point is at x = 400.

So, option b is correct.

Use the data from problem 5 to 8.

The total cost C(x) of a company as C(x) = 1000 + 25x + 2x² where x is the output.

Problem 5 :

Determine the average cost :

a)  1000/x + 25 + 2x       b)  1000/x + 20 + 2x

c)  1000/x + 30 + 3x       d)   1000/x + 25 + x

Solution :

Average cost = Total cost / output

= C(x) / x

= (1000 + 25x + 2x²) / x

= 1000/x + 25x/x + 2x² / x

= 1000/x + 25 + 2x

So, option a is correct.

Problem 6 :

Determine the marginal cost 

a)  30 + 4x     b)  25 + 4x      c)  50 + 4x    d) 50 + 5x

Solution :

While finding derivative of cost function, we get the marginal cost.

C(x) = 1000 + 25x + 2x²

d(C(x)) = d(1000 + 25x + 2x²)

= 0 + 25(1) + 2(2x)

marginal cost = 25 + 4x

So, option b is correct.

Problem 7 :

Find the marginal cost when 15 units are produced,

a)  60   b)  90     c)  80     d)  85

Solution :

Marginal cost = 25 + 4x

When x = 15

= 25 + 4(15)

= 25 + 60

= 85

So, option d is correct.

Problem 8 :

Find the actual cost of producing 15th unit.

a)  80    b)  70      c)  83        d)  90

Solution :

C(x) = 1000 + 25x + 2x²

To find actual cost of producing 15th unit, we have to find the difference between the cost of producing 15th unit and 14th unit.

Applying x = 15 and 14

C(15) - C(14)

= [1000 + 25(15) + 2(15)²] - [1000 + 25(14) + 2(14)²]

= 375 + 450 - 350 - 392

= 825 - 742

= 83

So, cost of producing 15th units is 83.

Use the data from problem 9 to 12.

The total cost function of a firm is given as

C(x) = 0.002x³ - 0.04x² + 5x + 1500

Where x is the output.

Problem 9 :

Determine the average cost 

a) 0.002x² - 0.04x + 5 + 1500/x

b) 0.002x² - 0.05x + 5 + 1500/x

c) 0.002x² - 0.05x + 5 + 1000/x

d) 0.002x² - 0.05x + 5 + 500/x

Solution :

Average cost = Total cost / output

C(x) = 0.002x³ - 0.04x² + 5x + 1500

= [0.002x³ - 0.04x² + 5x + 1500]/x

= 0.002x² - 0.04x + 5 + (1500/x)

So, the average cost is  0.002x² - 0.04x + 5 + (1500/x)

So, option a is correct.

Problem 10 :

Determine the marginal average cost (MAC)

a)  0.004x - 0.08 - 1500/x²

b)  0.004x - 0.04 - 1500/x²

c)  0.004x - 0.04 - 1000/x²

d)  0.001x - 0.04 - 1500/x²

Solution :

By finding the derivative of average cost, we will get marginal average cost.

AC = 0.002x² - 0.04x + 5 + (1500/x)

d(AC) = d(0.002x² - 0.04x + 5 + (1500/x))

= d(0.002x² - 0.04x + 5 + 1500x^-1)

= 0.002(2x) - 0.04(1) + 0 - 1500/x²)

= 0.004x - 0.04 - 1500/x²

So, option b is correct.

Problem 11 :

Find the marginal cost

a)   0.006x² - 0.10x + 5     b)  0.006x² - 0.16x + 5

c)   0.006x² - 0.08x + 5     b)  0.005x² - 0.08x + 5

Solution :

By finding the cost function, we will get marginal cost.

C(x) = 0.002x³ - 0.04x² + 5x + 1500

d(C(x)) = d(0.002x³ - 0.04x² + 5x + 1500)

= 0.002(3x²) - 0.04(2x) + 5(1) + 0

= 0.006x² - 0.08x + 5

So, option c is correct.

Problem 12 :

Find the rate of change MC with respect to x.

a)   0.012x - 0.10      b) 0.010x - 0.08

c)  0.012x + 0.08      d)  0.012x - 0.08

Solution :

To find the rate of change of MC, we have to find the derivative of marginal cost.

MC = 0.006x² - 0.08x + 5

d(marginal cost) = d(0.006x² - 0.08x + 5)

= 0.006(2x) - 0.08(1) + 0

= 0.012x - 0.08

So, the rate of change of marginal cost is 0.012x - 0.08