We start by choosing our variables.

Let x=The number of hours per week Niki will work at Job I.x=The number of hours per week Niki will work at Job I. size 12{x="The number of hours per week Niki will work at Job I" "." } {}

and y=The number of hours per week Niki will work at Job II.y=The number of hours per week Niki will work at Job II. size 12{y="The number of hours per week Niki will work at Job II" "." } {}

Now we write the objective function. Since Niki gets paid $40 an hour at Job I, and $30 an hour at Job II, her total income I is given by the following equation.

Our next task is to find the constraints. The second sentence in the problem states, "She never wants to work more than a total of 12 hours a week." This translates into the following constraint:

The third sentence states, "For every hour she works at Job I, she needs 2 hours of preparation time, and for every hour she works at Job II, she needs one hour of preparation time, and she cannot spend more than 16 hours for preparation." The translation follows.

The fact that xx size 12{x} {} and yy size 12{y} {} can never be negative is represented by the following two constraints:

x≥0x≥0 size 12{x >= 0} {}, and y≥0y≥0 size 12{y >= 0} {}.

Well, good news! We have formulated the problem. We restate it as

Maximize I = 40 x + 30 y I = 40 x + 30 y size 12{I="40"x+"30"y} {}

Subject to: x + y ≤ 12 x + y ≤ 12 size 12{x+y <= "12"} {}

In order to solve the problem, we graph the constraints as follows.

Figure 1

Observe that we have shaded the region where all conditions are satisfied. This region is called the feasibility region or the feasibility polygon.

The Fundamental Theorem of Linear Programming states that the maximum (or minimum) value of the objective function always takes place at the vertices of the feasibility region.

Therefore, we will identify all the vertices of the feasibility region. We call these points critical points. They are listed as (0, 0), (0, 12), (4, 8), (8, 0). To maximize Niki's income, we will substitute these points in the objective function to see which point gives us the highest income per week. We list the results below.

Clearly, the point (4, 8) gives the most profit: $400.

Therefore, we conclude that Niki should work 4 hours at Job I, and 8 hours at Job II.

We choose our variables.

Let x=The number of regular gadgets manufactured each day.x=The number of regular gadgets manufactured each day. size 12{x="The number of regular gadgets manufactured each day" "." } {}

and y=The number of premium gadgets manufactured each day.y=The number of premium gadgets manufactured each day. size 12{y="The number of premium gadgets manufactured each day" "." } {}

The objective function is

We now write the constraints. The fourth sentence states that the company can make at most 7 gadgets a day. This translates as

Since the regular gadget requires one hour of assembly and the premium gadget requires two hours of assembly, and there are at most 12 hours available for this operation, we get

Similarly, the regular gadget requires two hours of finishing and the premium gadget one hour. Again, there are at most 12 hours available for finishing. This gives us the following constraint.

The fact that xx size 12{x} {} and yy size 12{y} {} can never be negative is represented by the following two constraints:

x≥0x≥0 size 12{x >= 0} {}, and y≥0y≥0 size 12{y >= 0} {}.

We have formulated the problem as follows:

Maximize P = 20 x + 30 y P = 20 x + 30 y size 12{P="20"x+"30"y} {}

Subject to: x + y ≤ 7 x + y ≤ 7 size 12{x+y <= 7} {}

In order to solve the problem, we graph the constraints as follows:

Figure 2

Again, we have shaded the feasibility region, where all constraints are satisfied.

Since the extreme value of the objective function always takes place at the vertices of the feasibility region, we identify all the critical points. They are listed as (0, 0), (0, 6), (2, 5), (5, 2), and (6, 0). To maximize profit, we will substitute these points in the objective function to see which point gives us the maximum profit each day. The results are listed below.

The point (2, 5) gives the most profit, and that profit is $190. Therefore, we conclude that we should manufacture 2 regular gadgets and 5 premium gadgets daily for a profit of $190.

The graph is shown below.

Figure 3

The five critical points are listed in the above figure. Clearly, the point (2, 2) maximizes the objective function to a maximum value of 50. The reader should observe that the first constraint x+y≥1x+y≥1 requires that feasibility region must be bounded below by the line x+y=1x+y=1.

We choose the variables as follows:

Let x=The number of hours per week John is employed.x=The number of hours per week John is employed. size 12{x="The number of hours per week John is employed" "." } {}

and y=The number of hours per week Mary is employed.y=The number of hours per week Mary is employed. size 12{y="The number of hours per week Mary is employed" "." } {}

The objective function is

The fact that each student must work at least one hour each week results in the following two constraints:

Since John can grade 20 papers per hour and Mary 30 papers per hour, and there are at least 110 papers to be graded per week, we get

The fact that x and y are non-negative, we get

x≥0x≥0 size 12{x >= 0} {}, and y≥0y≥0 size 12{y >= 0} {}.

The problem has been formulated as follows.

Minimize C = 5x + 8y C = 5x + 8y size 12{C=5x+8y} {}

Subject to: x ≥ 1 x ≥ 1 size 12{x >= 1} {}

To solve the problem, we graph the constraints as follows:

Figure 5

Again, we have shaded the feasibility region, where all constraints are satisfied.

Since the extreme value of the objective function always takes place at the vertices of the feasibility region, we identify the two critical points, (1, 3) and (4, 1). To minimize cost, we will substitute these points in the objective function to see which point gives us the minimum cost each week. The results are listed below.

The point (4, 1) gives the least cost, and that cost is $28. Therefore, we conclude that Professor Symons should employ John 4 hours a week, and Mary 1 hour a week at a cost of $28 per week.

We choose the variables as follows.

Let x = The number of days Mr. Hamer eats Pasta.

and y = The number of days Mr. Hamer eats Tofu.

Since he is trying to minimize his cholesterol intake, our objective function represents the total amount of cholesterol C provided by both meals.

The constraint associated with the total amount of protein provided by both meals is as follows:

Similarly, the two constraints associated with the total amount of carbohydrates and vitamins are obtained, and they are

The constraints that state that x and y are non-negative are

We summarize all information as follows:

Minimize C=60x+50yC=60x+50y

Subject to: 8x+16y≥2008x+16y≥200

To solve the problem, we graph the constraints as follows.

Figure 6

Again, we have shaded the unbounded feasibility region, where all constraints are satisfied.

To minimize the objective function, we find the vertices of the feasibility region. These vertices are (0, 24), (8, 12), (15, 5) and (25, 0). To minimize cholesterol, we will substitute these points in the objective function to see which point gives us the smallest value. The results are listed below.

The point (8, 12) gives the least cholesterol, which is 1080 mg. This states that for every 20 meals, Professor Hamer should eat Pasta 8 days, and Tofu 12 days.