You would like to make a nutritious meal of eggs, edamame, and elbow macaroni. the meal should provide at least 40 g of carbohydrates, at least 20 g of protein, and no more than 50 g of fat. an egg (one serving) contains 2g of carbohydrates, 17 g of protein, and 14 g of fat. a serving of edamame contains 12 g of carbohydrates, 12 g of protein, and 6 g of fat. a serving of elbow macaroni contains 43 g of carbohydrates, 8 g of protein, and 1 g of fat. an egg costs $2, a serving of edamame costs $ 5, and a serving of elbow macaroni costs $3. formulate a linear optimization model that could be used to determine the number of servings of egg, edamame, and elbow macaroni that should be in the meal in order to meet the nutrition requirements at minimal cost. (you don't need to find the optimal solution to the model you formulate.)

Lets denote

1. eggs as x,

2. edamame as y

3. elbow macaroni as z

The problem then is

min TC=2x+5y+3z

subject to

Step-by-step explanation:

First the objective is to minimize total cost subject to some nutritional requirements.

So the total cost function (TC) is the number of servings multiplied for the corresponding costs. Eggs cost 2, edamame 5, and macaroni 3

Next we have to meet the nutritional requirements, the first of the restrictions is the protein restriction. The problem requires that the meal contains at least 40g of carbohydrates (that is why the restriction is ). Then we add how much each meal component adds to the total, eggs add 2g of carbs, edamame 12g, and macaroni 43g.

Same for protein, we need at least 20 grams of protein (). Eggs add 17g, edamame adds 12g, and macaroni adds 8g.

Finally we don't want more than 50 grams of fat (). Eggs add 14g, edamame add 6g and macaroni 1g.

Finally, we add the non negativity restrictions-> we cannot buy negative quantities of these goods, but we allow for zero.

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explanation

the graph has a slope of 1/3 because each time you move up 1, you go to the right 3. so, slope = rise/run = 1/3. we can cross choice a off the list since we want a slope of 3, rather than 1/3.

the first table shown has a slope of -3. this is because the rise is -6 and the run is 2, so slope = rise/run = -6/2 = -3. note how x increases by 2 each time (run) and y drops by 6 each time (rise). choice b can be eliminated.

the second table has a slope of 3. each time x goes up by 1, the y values increase by 3. slope = rise/run = 3/1 = 3. alternatively you can use the slope formula to prove that the slope is 3 here. use any two rows of values to generate the two points. choice c is one of the answers

the third table also has a slope of 3. let's use the slope formula to prove as such. i'm going to use the first two rows of the table. so i'll use the two points (2,1) and (5,10)

m = slope

m = (y2-y1)/(x2-x1)

m = (10-1)/(5-2)

m = 9/3

m = 3 so we get a slope of 3

choice d is the other answer