# Linear Programming Worksheet - Page 2

Linear Programming Worksheet
• Page 2
11.
Find the point at which the maximum value of P = 10$x$ + 5$y$ subject to the constraints $x$ ≥ 0, $y$ ≥ 0 and 3$x$ + 4$y$ ≤ 12 occurs.
 a. (4, 0) b. (0, 0) c. (0, 3) d. (1, 1)

#### Solution:

Objective function is P = 10x + 5y

Constraints are x ≥ 0, y ≥ 0 and 3x + 4y ≤ 12.

The feasible region determined by the constraints is shown.
[Draw the constraints.]

From the graph the vertices are (0, 0), (4, 0) and (0, 3).

To evaluate the minimum and maximum values of P, we evaluate P = 10x + 5y at each of the vertices.

At (0, 0), P = 10(0) + 5(0) = 0
[Substitute the values.]

At (4, 0), P = 10(4) + 5(0) = 40
[Substitute the values.]

At (0, 3), P = 10(0) + 5(3) = 15
[Substitute the values.]

So, the maximum value of P is 40 which occurs when x = 4 and y = 0 .

12.
Find the values of $x$ and $y$ that maximize the objective function P = 7$x$ + 8$y$ subject to the constraints $x$ ≥ 0, $y$ ≥ 0, $x$ - $y$ ≤ 1 and $x$ + $y$ ≤ 3.
 a. $x$ = 2, $y$ = 1 b. $x$ = 4, $y$ = 5 c. $x$ = 1, $y$ = 2 d. $x$ = 0, $y$ = 3

#### Solution:

Objective function is P = 7x + 8y

Constraints are x ≥ 0, y ≥ 0, x - y ≤ 1 and x + y ≤ 3.

The feasible region determined by the given constraints is shown.

From the graph the vertices are (0, 0), (1, 0), (2, 1) and (0, 3)

To find the minimum and maximum values of P, we evaluate P = 7x + 8y at each of the four vertices.

At (0, 0), P = 7(0) + 8(0) = 0
[Substitute the values.]

At (1, 0), P = 7(1) + 8(0) = 7
[Substitute the values.]

At (2, 1), P = 7(2) + 8(1) = 22
[Substitute the values.]

At (0, 3), P = 7(0) + 8(3) = 24
[Substitute the values.]

So, the maximum value of P is 24 which occurs when x = 0 and y = 3.

13.
What are the values of $x$ and $y$ that minimize the objective function D = 7$x$ + 2$y$ subject to the constraints $x$ ≥ 0, $y$ ≥ 0, $x$ ≤ 2$y$ + 1 and $y$ ≤ 4 - $x$ respectively?
 a. $x$ = 1, $y$ = 0 b. $x$ = 0, $y$ = 0 c. $x$ = 0, $y$ = 1 d. $x$ = 0, $y$ = 4

#### Solution:

Objective function is D = 7x + 2y

Constraints are x ≥ 0, y ≥ 0, x ≤ 2y + 1 and y ≤ 4 - x

The feasible region determined by the given constraints is shown.

The vertices are (0, 0), (1, 0), (3, 1), (0, 4)

To find the minimum and maximum values of D, we evaluate D = 7x + 2y at each of the four points.

At (0, 0), D = 7(0) + 2(0) = 0
[Substitute the values.]

At (1, 0), D = 7(1) + 2(0) = 7
[Substitute the values.]

At (3, 1), D = 7(3) + 2(1) = 23
[Substitute the values.]

At (0, 4), D = 7(0) + 2(4) = 8
[Substitute the values.]

So, the minimum value of D is Ã¢â‚¬Ëœ0Ã¢â‚¬â„¢ which occurs when x = 0, y = 0.

14.
A factory produces two types of toys A and B with the help of two machines I and II. The two machines I and II can not be used for not more than $1\frac{1}{2}$ hours each to manufacture the toys. The toy A requires 1 minute of machine I and 2 minutes of II, where as a toy B takes 3 minutes of I and 1 minute of II. The profit for A is $90 and for B$80. How many of each type of toys are to be produced to give the maximum profit?
 a. 80 toys of A, 70 toys of B b. 45 toys of A, 30 toys of B c. 60 toys of A, 10 toys of B d. 36 toys of A, 18 toys of B

#### Solution:

Let the number of toys A and B to be made be x and y respectively.

The objective function is the profit P = 90x + 80y.

To make x number of toys A and y number of toys B, the machine I should be used for 1(x) + 3(y) minutes. But one cannot use the machine I for more than 11 / 2 hours, we have x + 3y ≤ 90
[ 11 / 2 hour = 90 minutes.]

To make x number of toys A and y number of toys B, the machine II should be used for 2(x) + 1(y) minutes. But one cannot use the machine y for more than 11 / 2 hours, we have 2x + y ≤ 90
[11 / 2 hour = 90 minutes.]

x ≥ 0, y ≥ 0
[x , y are number of toys.]

The feasible region determined by the above constraints is shown.

From the figure the vertices are (0, 0), (45, 0), (36, 18) and (0, 30).

At (0, 0), P = 90(0) + 80(0) = 0.
[Substitute the values.]

At (45, 0), P = 90(45) + 80(0) = 4050.
[Substitute the values.]

At (36, 18), P = 90(36) + 80(18) = 4680.
[Substitute the values.]

At (0, 30), P = 90(0) + 80(30) = 2400.
[Substitute the values.]

So, the maximum profit can be obtained by making 36 toys of A and 18 toys of B.

15.
What are the minimum and maximum values of the objective function P = 5$x$ + 6$y$ subject to the constraints $x$ ≥ 2, $x$ ≤ 4, $y$ ≥ 1, $y$ ≤ 3?
 a. 28, 38 b. 26, 28 c. 16, 38 d. 16, 28

#### Solution:

Objective function is P = 5x + 6y

Constraints are x ≥ 2, x ≤ 4, y ≥ 1, y ≤ 3

The feasible region determined by the above constraints is shown.

From the figure the vertices are (2, 1), (4, 1), (2, 3) and (4, 3).

At (2, 1), P = 5(2) + 6(1) = 16
[Substitute the values.]

At (4, 1), P = 5(4) + 6(1) = 26
[Substitute the values.]

At (2, 3), P = 5(2) + 6(3) = 28
[Substitute the values.]

At (4, 3), P = 5(4) + 6(3) = 38
[Substitute the values.]

So, the minimum and maximum values of P are 16 and 38.

16.
A doctor wishes to mix together two kinds of food, X and Y, in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one pound of food are shown. One pound of food X costs $4 and one pound of food Y costs$5. Find what combination of foods should be used to have the least cost?

 a. mixing 8 pounds of food X with 0 pounds of food Y b. mixing 2 pounds of food X with 4 pounds of food Y c. mixing 2 pounds of food X with 3 pounds of food Y d. mixing 10 pounds of food X with 0 pounds of food Y

#### Solution:

Let x pounds of food X with y pounds of food Y to be mixed so that it contains 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C.

Since 1 pound of food X is $4 and 1 pound of food Y is$5, we have to minimize 4x + 5y, So objective function is z = 4x + 5y

From the given table the constraints are x + 2y ≥ 10, 2x + 2y ≥ 12 or x + y ≥ 6 and 3x + y ≥ 8

x ≥ 0, y ≥ 0
[x, y are number of pounds.]

The feasible region determined by the above constraints is shown.

From the figure the vertices are (10, 0), (2, 4), (1, 5) and (0, 8)

At (10, 0), Z = 4(10) + 5(0) = 40
[Substitute the values.]

At (2, 4), Z = 4(2) + 5(4) = 28
[Substitute the values.]

At (1, 5), Z = 4(1) + 5(5) = 29
[Substitute the values.]

At (0, 8), Z = 4(0) + 5(8) = 40
[Substitute the values.]

Since the point (2, 4) is giving the minimum value of Z, mixture of least cost consists 2 pounds of food X and 4 pounds of food Y.

17.
Which graphical tool identifies conditions that make as large as possible or as small as possible?
 a. Linear programming b. Graph theory c. Curve sketching d. Optimization technique

#### Solution:

The graphical tool, which identifies conditions that make as large as possible or as small as possible is known as linear programming.
[Definition.]

18.
The quantity to be maximized or minimized is represented by a linear function called ______.
 a. The restriction b. The constraint c. The feasible function d. The objective function

#### Solution:

The quantity to be maximized or minimized is represented by a linear function called the objective function.
[Definition.]

19.
The graph of the linear system of constraints is called ______.
 a. The constraint region b. The feasible region c. The linear region d. The linear graph

#### Solution:

The graph of the linear system of constraints is called the feasible region.
[Definition.]

20.
Limits on the variables in the objective function are called ____.
 a. Restrictions b. Objectives c. Vertices d. Extremes

#### Solution:

Limits on the variables in the objective function are called restrictions.
[Definition.]