High School

You are a coffee enthusiast. Every morning, the minute you get up, you make yourself some pour-over coffee in your Chemex. You are meticulous, weighing the coffee beans and the water, measuring the water's temperature, and timing the pour to achieve optimal results. You buy your beans at Northampton Coffee, where a 12 oz bag costs you $16.95. Although you would prefer to use bottled water to make the best coffee possible, you are environmentally conscious and thus use Northampton tap water, which costs $5.72 for every 100 cubic feet.

You find your coffee tastes equally good as long as you have anywhere between 16 to 17 grams of water for each gram of coffee beans. You want to have between 350 and 380 milliliters of coffee (i.e., water) to start your day right. You use an additional 250 milliliters of boiling water to "wash" the filter and warm the Chemex and your cup. You use one filter every morning, which you buy in packs of 100 for $18.33.

You heat your water with a 1 kW electric kettle, which takes 5 minutes to reach the desired temperature. Your 1.5 kW grinder takes 30 seconds to grind the coffee beans. Through National Grid, you pay $0.11643 for each kWh you use (i.e., this would be the cost of running the kettle for a full hour or running the grinder for 40 minutes).

(a) What ratio of water to beans and what quantity of coffee do you think will minimize the cost of your morning coffee? Why? (You don't need to calculate anything now.)

(b) Actually calculate the minimum cost of your daily coffee-making process. (In this moment, you might curse the fact that you live in a place that uses the imperial system. One ounce is roughly 28.3495 grams, and one foot is 30.48 centimeters. In the metric system, you can assume that one gram of water is equal to one milliliter of water, which is equal to one cubic centimeter of water.)

(c) Now calculate the maximum cost of your daily coffee-making process.

(d) Reformulate what you did in (b) and (c) in terms of what you learned in linear algebra: determine what your variables are, write what the constraints are, and what the objective function is (i.e., the function that you are maximizing or minimizing).

(e) Graph the constraints you found in (d) - this gives you the feasible region.

(f) How could you have found the answers to (b) and (c) with the picture you drew in (e)? What does 'minimizing' or 'maximizing' a function over your feasible region mean? How can you find the optimal solution(s)? You might have seen this in high school as the graphical method. If you haven't, plot on your graph the points where your objective function evaluates to 0. Then do the same for 1. What do you notice?

(g) How expensive would Northampton's water have to become so that the cheaper option becomes a different ratio of water to beans than the one you found in (a)?

(h) Now suppose that instead of maximizing or minimizing the cost of your coffee-making process, you are minimizing \(\alpha c + \beta w\) where \(c\) is the number of grams of coffee beans you use and \(w\) is the number of grams of water you use, and \(\alpha, \beta \in \mathbb{R}\). What are the potential optimal solutions? Can any point in your feasible region be an optimal solution? Why or why not?

(i) For each potential optimal solution in (h), characterize fully for which pairs \((\alpha, \beta)\) the objective function \(\alpha c + \beta w\) is minimized on that particular optimal solution. (If you're not sure how to start, try different values of \(\alpha\) and \(\beta\) and find where \(\alpha c + \beta w\) is minimized.)

(j) Can you state what happens in (i) more generally and prove it?

Answer :

a) The ratio of water to beans that will minimize the cost of morning coffee is 17:1, while the quantity of coffee is 17 grams.

b) The following is the calculation of the minimum cost of your daily coffee-making process:

$ / day = (16.95 / 12 * 17) + (5.72 / 100 * 0.17) + (18.33 / 100) + (0.11643 / 60 * (5/60 + 0.5)) = 1.413 dollars.

c) The following is the calculation of the maximum cost of your daily coffee-making process:

$ / day = (16.95 / 12 * 16) + (5.72 / 100 * 0.16) + (18.33 / 100) + (0.11643 / 60 * (5/60 + 0.5)) = 1.413 dollars.

d) Variables: amount of coffee beans (c), amount of water (w)

Constraints: 16 ≤ c ≤ 17; 350 ≤ w ≤ 380;

w = 17c

Objective Function: 16.95/12c + 5.72w/100 + 18.33/100 + (0.11643 / 60 * (5/60 + 0.5))

e) Constraints: 16 ≤ c ≤ 17; 350 ≤ w ≤ 380; w = 17c,

graph shown below:

f) The optimal solution(s) can be found at the vertices of the feasible region. Minimizing or maximizing a function over the feasible region means finding the highest or lowest value that the function can take within that region. The optimal solution(s) can be found by evaluating the objective function at each vertex and choosing the one with the lowest value. The minimum value of the objective function is found at the vertex (16, 272) and is 1.4125 dollars. The maximum value of the objective function is found at the vertex (17, 289) and is 1.4375 dollars.

g) The cost of Northampton's water would have to increase to $0.05 per 100 cubic feet for the cheaper option to become a different ratio of water to beans.

h) The potential optimal solutions are all the vertices of the feasible region. Any point in the feasible region cannot be an optimal solution because the objective function takes on different values at different points.

i) The potential optimal solutions are:(16, 272) for α ≤ 0 and β ≥ 0(17, 289) for α ≥ 16.95/12 and β ≤ 0

All other points in the feasible region are not optimal solutions.

ii) The objective function αc + βw is minimized for a particular optimal solution when α is less than or equal to the slope of the objective function at that point and β is greater than or equal to zero.

This is because the slope of the objective function gives the rate of change of the function with respect to c, while β is a scaling factor for the rate of change of the function with respect to w.

Learn more about Constraints:

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