Question

LED bulbs are energy-efficient because they use significantly less electricity than traditional bulbs while producing the same amount of light. They convert more energy into light rather than heat, reducing waste. Additionally, their long lifespan means fewer replacements, saving resources and money over time. A company manufactures a new type of energy-efficient LED bulb. The cost of production and the revenue generated by selling x bulbs (in an hour) are modelled as C(x) = 0.5x2 − 10x + 150 and R(x) = −0.3x2 + 20x respectively, where C(x) and R(x) are both in ₹.

To maximize the profit, the company needs to analyze these functions using calculus. Use the given models to answer the following questions:

1. Derive the profit function P(x).   (1)
2. Find the critical points of P(x).   (1)
3. 1. Determine whether the critical points correspond to a maximum or a minimum profit by using the second derivative test.  (2)
      OR
   2. Identify the possible practical value of x (i.e., the number of bulbs that can realistically be produced and sold) that can maximize the profit if the resources available and the expenditure on machines allows to produce minimum 10 but not more than 18 bulbs per hour. Also calculate the maximum profit.

Answer 1

Given:

Cost = C(x) = 0.5x² − 10x + 150

Revenue = R(x) = −0.3x² + 20x

I. Let P(x) be the profit.

Now, Profit = Revenue − Cost

P(x) = R(x) – C(x) 

= –0.3x² + 20x – (0.5x² − 10x + 150)

= –0.3x² + 20x – 0.5x² + 10x – 150

= (–0.5 – 0.3)x² + 20x + 10x – 150

= –0.8x² + 30x – 150

II. To find critical points, we find values of x where P'(x) = 0

Now, P(x) = –0.8x² + 30x – 150

Differentiate with respect to x.

P'(x) = `(d(-0.8x^2 + 30x - 150))/dx`

= –0.8 × 2x + 30

= –1.6x + 30

Putting P'(x) = 0

–1.6x + 30 = 0

–1.6x = –30

x = `30/1.6`

x = `30/(16/10)`

x = `30/16 xx 10`

x = 18.75

Thus, 18.75 is the critical point of P(x).

III. A. P(x) = = –0.8x² + 30x – 150

P'(x) = –1.6x + 30

We need to differentiate again for the second derivative test.

Differentiate P'(x) with respect to x.

P''(x) = `(d(-1.6 x + 30))/(dx)`

P''(x) = –1.6

Since P''(x) < 0,

x = 18.75 is the maxima.

So, we can write x = 18.75 as the critical point corresponding to maximum profit.

OR

III. B. Now, P(x) = –0.8x² + 30x – 150

It says the value of x is a minimum of 10 and not more than 18.

So, 10 ≤ x ≤ 18

We find P(10) and P(18) and check what is maximum.

Finding P(10):

P(x) = –0.8x² + 30x – 150

= –0.8 × 10² + 30 × 10 – 150

= –0.8 × 100 + 300 – 150

= –80 + 300 – 150

= 70

Finding P(18):

P(x) = –0.8x² + 30x – 150

P(18) = –0.8 × 18² + 30 × 18 – 150

= –0.8 × 324 + 540 – 150

= –259.2 + 390

= 130.8

Thus, maximum profit occurs when x is 18, i.e. 18 bulbs and maximum profit is ₹ 130.8.