Consider an underground aquifer where the initial stock is X0=100,000 acre-feet. Andres bought the exclusive rights to pump water from the aquifer over a period of 20 time periods from the government. The revenues in a time period, t, is given by Rt = aWt – (b/2) W 2 where Wt is the amount of water pumped in acre-feet. The cost of pumping water is Ct = cWt/Xt. The stock of water in the aquifer in the future depends on the current stock minus the current amount of water pumped plus a constant recharge rate equal to D during each period. Q1: Write the objective function along with the constraints if Andres tries to maximize profit over his planning horizon. Assume that Andres does not value any leftover stock after his rights to pumping water expires and his discount rate is begin mathsize 12px style delta end style or, equivalently, his discount factor is ρ.
Q2: Assume that a=100, b=0.001, c=812,500, D = 35,000 and begin mathsize 12px style delta end style = 0.05. In an excel worksheet (sheet3), set up the problem above using an initial guess of Wt=100 for all relevant periods. Copy everything to sheet4 and use solver to derive the optimal water extraction path that maximizes profit. Graph extraction over time and stock over time in separate figures.

i don't see nikolas work.

step-by-step explanation:

10 * 563.67 = 5636.7

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation:

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation: