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.

answer: 12

step-by-step explanation:

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3