Question

Mr. Avinash manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is ` 30. On the other hand, the cost, in rupees, of producing x units is $240 + bx + cx^2$ , where b and c are some constants. Avinash noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.66 percent. However, an increase in daily production from 40 to 60 units results in an increase of only 50 percent in the daily production cost. Assume that demand is unlimited and that Avinash can sell as much as he can produce. His objective is to maximize the profit. How many units should Mr. Avinash produce daily?

• A. 70
• B. 100
• C. 130
• D. 150
• E. 180

Answer 1

The Correct Option is B

The correct answer is option (B):

100

Let's break down this profit maximization problem step by step.

First, we need to understand the cost function: `C(x) = 240 + bx + cx^2`. We are also given that the selling price per unit is `30`. The revenue from selling x units is `R(x) = 30x`. The profit function, `P(x)`, is the difference between revenue and cost: `P(x) = R(x) - C(x) = 30x - (240 + bx + cx^2) = 30x - 240 - bx - cx^2`.

Now let's use the information about the production cost changes.

* Doubling production from 20 to 40 units increases the cost by 66.66% (which is approximately 2/3).

`C(40) - C(20) = (2/3) * C(20)`
`C(40) = (5/3) * C(20)`

Substituting the cost function:
`240 + 40b + 1600c = (5/3) * (240 + 20b + 400c)`
`720 + 120b + 4800c = 1200 + 100b + 2000c`
`20b + 2800c = 480` (Equation 1)

* Increasing production from 40 to 60 units increases the cost by 50% (which is 1/2).

`C(60) - C(40) = (1/2) * C(40)`
`C(60) = (3/2) * C(40)`

Substituting the cost function:
`240 + 60b + 3600c = (3/2) * (240 + 40b + 1600c)`
`480 + 120b + 7200c = 720 + 120b + 4800c`
`2400c = 240`
`c = 1/10 = 0.1`

Substitute `c = 0.1` into Equation 1:
`20b + 2800 * 0.1 = 480`
`20b + 280 = 480`
`20b = 200`
`b = 10`

So, the cost function is `C(x) = 240 + 10x + 0.1x^2`. Now we can refine our profit function:
`P(x) = 30x - 240 - 10x - 0.1x^2 = 20x - 240 - 0.1x^2`

To maximize profit, we need to find the critical points by taking the derivative of the profit function with respect to x and setting it equal to zero:
`P'(x) = 20 - 0.2x = 0`
`0.2x = 20`
`x = 100`

To verify that this is a maximum, we can take the second derivative:
`P''(x) = -0.2`
Since the second derivative is negative, we have a maximum at x = 100.

Therefore, Mr. Avinash should produce 100 units daily to maximize his profit.