Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producingxunits is 240 +bx+cx2, wherebandcare some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.66%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.What is the maximum daily profit, in rupees, that Mr. David can realize from his business?a)620b)920c)840d)760e)cannot be determinedCorrect answer is option 'D'. Can you explain this answer?

Question

Answer

To maximize daily profit, Mr. David needs to find the optimal number of units to produce. Let's analyze the given information step by step to determine the maximum daily profit.

1. Cost Function:
The cost of producing x units is given by the equation: Cost(x) = 240 + b*x + c*x^2, where b and c are constants.

2. Increase in Daily Production:
Doubling the daily production from 20 to 40 units increases the daily production cost by 66.66%. This means that the cost of producing 40 units is 1.6666 times the cost of producing 20 units.

3. Cost Increase Calculation:
Let's assume the cost of producing 20 units is C1, and the cost of producing 40 units is C2.
According to the given information, we have:
C2 = C1 + 66.66% of C1
C2 = C1 + 0.6666*C1
C2 = 1.6666*C1

4. Second Increase in Daily Production:
An increase in daily production from 40 to 60 units results in a 50% increase in the daily production cost. This means that the cost of producing 60 units is 1.5 times the cost of producing 40 units.

5. Cost Increase Calculation:
Let's assume the cost of producing 40 units is C2, and the cost of producing 60 units is C3.
According to the given information, we have:
C3 = C2 + 50% of C2
C3 = C2 + 0.5*C2
C3 = 1.5*C2

6. Profit Calculation:
Profit is calculated by subtracting the cost from the selling price per unit.
Profit(x) = Revenue(x) - Cost(x)
Profit(x) = x*30 - (240 + b*x + c*x^2)

7. Maximum Daily Profit:
To find the maximum daily profit, we need to determine the optimal number of units to produce. This can be done by analyzing the profit function and finding its maximum value.

8. Calculating Optimal Number of Units:
To find the optimal number of units, we need to differentiate the profit function with respect to x and set it equal to zero.
Profit'(x) = 30 - (b + 2c*x) = 0
b + 2c*x = 30
2c*x = 30 - b
x = (30 - b) / (2c)

9. Substituting Values:
We don't have specific values for b and c, so we cannot determine the exact optimal number of units. However, we can calculate the profit for different values of x and find the maximum daily profit.

10. Answer:
The maximum daily profit, in rupees, that Mr. David can realize from his business is given by the profit function at the optimal number of units. Since we cannot determine the exact optimal number of units without specific values for b and c, the answer cannot be determined.

Answer

Cost price function f(x) = cx² + bx + 240.
When x = 20, f(20) = 400c + 20b + 240
f(40) = 1600c + 40b + 240
66.67% is 2/3.
f(40) – f(20) = 2/3 * f(20)
Substituting f(40) and f(20), we get, 1200c+ 20b =(2/3) * (400c + 20b + 240)
Solving, we get 20b + 2800c = 480

Now, f(60) = 3600c + 60b + 240.
f(60) – f(40) = ½ * f(40).
Substituting, and solving, we get, 360 + 60b + 2400c = 240 + 60b + 3600c.
c = 1/10.

Substituting we get, b = 10.

So, the cost function f(x) = 0.1x² + 10x + 240.
Selling price = 30*x. ( Each unit SP is 30)
So, profit = SP – CP = 30x – (0.1x² + 10x + 240) = -x²/10 + 20x – 240.

For maximum profit, differentiating this should give 0.

dp/dt = 0, dp/dt = -x/5 + 20
=> 20 – x/5 = 0.

=>x= 100.

Also double differentiating d²p / dt², we get a negative number, so profit is maximum. So, profit is maximum when 100 units are produced daily.

The maximum profit can be calculated with the equation -x²/10 + 20x – 240, with x = 100, and it is 760.

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