Question

2. Wonder widgets Inc. projects that their monthly profit (in thousands) for producing widgets is given by P(x)-01x + 6x - 500. (a) How many widgets should they produce each month to maximize profit? (b) For this production, what is their monthly profit?

Answer 1

Given P(x) = -0.01x² + 6x - 500

(a)

Differentiating P(x) with respect to 'x' we get

P'(x) = (-0.01)(2x) + 6(1) - 0 {if f(x) = xⁿ then f'(x) = n*x^n-1}

P'(x) = -0.02x + 6

Differentiating P'(x) with respect to 'x' we get

P''(x) = (-0.02)(1) + 0 = -0.02

To find critical points where the function P(x) is maximum or minimum, equate P'(x) with zero and solve for 'x'

implies P'(x) = 0

implies -0.02x + 6 = 0

implies -0.02x = -6

implies 0.02x = 6

implies x = 6/0.02

implies x = 300

implies P(x) is minimum or maximum at x = 300

To find whether P(x) is minimum or maximum find P''(x) at critical points and

If P''(x) at a critical point is less than zero then the function is maximum at that point

If P''(x) at a critical point is greater than zero then the function is minimum at that point

Now P''(x) at x = 300 is P''(300) = -0.02

implies P''(x) at x = 300 is -0.02 which less than zero

implies P(x) is maximum at x = 300

Therefore 300 widgets should be produced each month to maximize profit

(b)

To find maximum profit i.e., monthly profit for above production, substitute x = 300 in P(x)

implies P(300) = -0.01(300²) + 6(300) - 500

implies P(300) = -0.01(90000) + 1800 - 500

implies P(300) = -900 + 1800 - 500

implies P(300) = 400

Therefore their monthly profit for x = 300 is 400 (in thousands)