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*xn-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)
2. Wonder widgets Inc. projects that their monthly profit (in thousands) for producing widgets is given...
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