Of Municipal Solid Waste Production - Math Problem Example

Model 2:

152 = a + b*20

217 = a + b*37

The solution is a=75.6, b=3.82. Therefore, the first linear function is:

Annual MSW = 75.6+3.82*(Number of years after 1960)

245. b) Model 1 predicts MSW in 2005 to be on the level of 245.6 mil tons (85.4+3.56*45=245.6).

Model 2 predicts MSW in 2005 to be on the level of 247.5 mil tons (75.6+3.82*45=247.5).

236. c) Both models have rather close predictions. In order to validate their predictive power it is necessary to check the predicted values vs. the actual values. For this purpose, I used the actual MSW data for 2003 provided by the United States Environmental Protection Agency (EPA)[1]. The actual MSW in 2003 was 236.2 mil tons.

The predicted results are 238.48 mil tons for Model 1 and 239.86 mil tons for Model 2. So Model 1 seems to be slightly better than Model 2 in predicting the actual level of MSW in the USA.

2. In order to find the maximum profit and the number of tonnages that must be processed to yield the maximum profit, I took the first derivative of the profit function and found its solution when it equals 0.
3. a) R(x) = 5x; C(x) = 0.0001x2 + 1.2x + 60.

Profit function is P(x) = 5x - 0.0001x2 - 1.2x - 60.

The derivative is P(x)´ = 5 - 0.0002x - 1.2 = 0.

X= 19,000; P= 36,040

The maximum profit is 36,040 and the number of tonnage needed to be processed is 19,000.

1. b) R(x) = 50x – 0.5x2; C(x) = 10x + 3.

Profit function is P(x) = 50x – 0.5x2 - 10x -3

The derivative is P(x)´ = 50 - x - 10 = 0.

X= 40; P= 797

The maximum profit is 797 and the number of tonnage needed to be processed is 40.

1. c) R(x) = 20x – 0.1x2; C(x) = 4x +2.

Profit function is P(x) = 20x - 0.1x2 - 4x -2.

The derivative is P(x)´ = 20 – 0.2x - 4 = 0.

X= 80; P= 638

The maximum profit is 638 and the number of tonnage needed to be processed is 80.