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.How many units should Mr. David produce daily ?a)130b)100c)70d)150e)cannot be determinedCorrect answer is option 'B'. Can you explain this answer?

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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 producing x units is 240 + bx + cx², where b and c are 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.

How many units should Mr. David produce daily ?

a)
130
b)
100
c)
70
d)
150
e)
cannot be determined

Correct answer is option 'B'. Can you explain this answer?

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Mr. David manufactures and sells a single product at a fixed price in ...

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.

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Mr. David manufactures and sells a single product at a fixed price in ...

Given information:
- Selling price of each unit = Rs. 30
- Cost of producing x units = 240 + bx + cx²
- Doubling the daily production from 20 to 40 units increases the daily production cost by 66.66%
- An increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost
- Demand is unlimited

To maximize profit, we need to find the optimum daily production:

Step 1: Calculate the daily production cost for different production levels:
Let's assume the daily production is x units.

For 20 units, the cost is 240 + b(20) + c(20)² = 240 + 20b + 400c = 240 + 20(b + 20c) = 240 + 20k (where k = b + 20c)

For 40 units, the cost is 240 + b(40) + c(40)² = 240 + 40b + 1600c = 240 + 40(b + 40c) = 240 + 40k (where k = b + 40c)

For 60 units, the cost is 240 + b(60) + c(60)² = 240 + 60b + 3600c = 240 + 60(b + 60c) = 240 + 60k (where k = b + 60c)

Step 2: Calculate the percentage increase in cost:
The percentage increase in cost from 20 units to 40 units is (240 + 40k - 240 - 20k)/ (240 + 20k) * 100% = 66.66%

The percentage increase in cost from 40 units to 60 units is (240 + 60k - 240 - 40k)/ (240 + 40k) * 100% = 50%

Step 3: Set up equations:
From step 2, we have two equations:
(240 + 40k - 240 - 20k)/ (240 + 20k) = 66.66%
(240 + 60k - 240 - 40k)/ (240 + 40k) = 50%

Simplifying the equations:
20k/ (240 + 20k) = 66.66%
20k/ (240 + 40k) = 50%

Step 4: Solve the equations:
Solving the first equation:
20k/ (240 + 20k) = 66.66%
20k = 0.6666(240 + 20k)
20k = 159.99 + 13.332k
6.668k = 159.99
k = 23.997

Solving the second equation:
20k/ (240 + 40k) = 50%
20k = 0.5(240 + 40k)
20k = 120 + 20k
0 = 120

Step 5: Determine the optimum daily production:
Since the second equation has no solution, it means that the increase in daily production from 40 units to 60 units

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