Mathematics 102 Solutions for HWK 5a Section p70 Problem 39. Use the Gauss-Jordan method to solve the following system of equations.

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1 Mathematics 102 Solutions for HWK 5a Section p70 Problem. Use the Gauss-Jordan method to solve the following system of equations. x + y z +2w = 20 2x y + z + w =11 x 2y + z 2w =27 Solution. The augmented matrix is Replace Row 2 by Row 2 2Row1. Replace Row by Row Row1. The new augmented matrix is Divide Row 2 by toobtain the new augmented matrix Replace Row 1 by Row 1 Row 2. Replace Row by Row + 5 Row 2. The new augmented matrix is Replace Row 2 by Row 2 Row to obtain Multiply the bottom row by 1 toobtain Page 1 of 6 A. Sontag February 1, 2004

2 Since x, y, and z each correspond to leading nonzero entries, use w as a parameter and solve for the others in terms of w. The first row of the augmented matrix tells us that x + w =, so x = w. The second row gives us y +4w = 1, so y = 1 4w. Finally, the bottom row gives us z +w = 2 soz = 2 w. The solutions are therefore where w can take any real-number value. (x, y, z, w) =( w, 1 4w, 2 w, w) 2.2 p70 Problem 51. Transportation A manufacturer purchases a part for use at both of its plants one at Roseville, California, the other at Akron, Ohio. The part is available in limited quantities from two suppliers. Each supplier has 75 units available. The Roseville plant needs 40 units, and the Akron plant requires 75 units. The first supplier charges $70 per unit delivered to Roseville and $0 delivered to Akron. Corresponding costs from the second supplier are $80 and $120, respectively. The manufacturer wants to order a total of 75 units from the first, less expensive supplier, with the remaining 40 units to come from the second supplier. If the company spends $10,750 to purchase the required number of units for the two plants, find the number of units that should be purchased from each supplier. Solution. Let x denote the number of units to be purchased from the first supplier to Roseville. Let y denote the number of units to be purchased from the first supplier to Akron. Let z denote the number of units to be purchased from the second supplier to Roseville. Let w denote the number of units to be purchased from the second supplier to Akron. Since Roseville needs 40 units, we must have x + z =40. Similarly, Akron needs 75 units, so y + w =75. All 75 of the units available from the first supplier will be purchased, so x + y =75. This leaves 40 units to be purchased from the second supplier, so z + w =40. Finally, since the total cost is $10,750, we must have 70x +0y +80z + 120w =10, 750. Using these equations in this order, the augmented matrix will be Page 2 of 6 A. Sontag February 1, 2004

3 Divide the bottom row by 10 to get Replace Row by Row Row 1. Replace Row 5 by Row 5 7Row1. The new augmented matrix is Replace Row by Row Row 2. Replace Row 5 by Row 5 Row 2, and the new matrix is Replace Row 4 by Row 4 + Row, then switch the new Row 4 with Row 5, then multiply Row by 1. The resulting matrix is Replace Row 1 by Row 1 Row and replace Row 4 by Row 4 Row to obtain Divide Row 4 by 2 and obtain Page of 6 A. Sontag February 1, 2004

4 Finally, replace Row 1 by Row 1 + Row 4, replace Row 2 by Row 2-Row4,andreplace Row by Row - Row 4. This gives us the augmented matrix This gives us x =40,y=5,z=0, and w = 40. In other words, the company should purchase 40 units from the first supplier for the Roseville plant and 5 units from the first supplier for the Akron plant, and it should purchase the remaining 40 units that are needed for the Akron plant from the second supplier, with none of the second supplier s units purchased for Roseville. 2.2 p70 Problem 57. Dietetics (reworded as described in a post on class conference) A hospital dietician is planning a special diet for a certain patient. The total amount per meal of food groups A, B, and C must equal 400 grams. The amount of food from group A is one third of the amount from group B, and the sum of the amounts of group A and group C should equal twice the amount of group B. (a) How many grams of each food group should be included? (b) Suppose we drop the requirement that the amount of group-a food is one third that of group-b food. Describe the set of all possible solutions. (c) Suppose that, in addition to the conditions given in the original problem, foods A and B cost 2 cents per gram and food C costs cents per gram, and that a meal must cust $8. Is a solution possible? Solution. (a) Let x, y, and z denote the number of grams of food groups A, B, and C, respectively to be used per meal. Since the total amount must be 400 grams we have x + y + z = 400. Since the amount of A is one third the amount of B, x = y or x y =0. Since the sum of the amounts of A and C must equal twice the amount of B, we also have x+z =2y or x 2y + z =0. The three displayed equations give us the augmented matrix (I m omitting the vertical divider this time) Replace Row 2 by Row 2 Row1and replace Row by Row Row 1. The new augmented matrix is Page 4 of 6 A. Sontag February 1, 2004

5 Multiply Row 2 by 1, multiply Row by 2, and then switch Rows 2 and. The augmented matrix becomes Replace Row 1 by Row 1 Row 2, and replace Row by Row 4Row 2toobtain Replace Row 1 by Row 1 Row to obtain Finally, divide Row 1 by, Row 2 by, and Row by. This gives us the augmented matrix Converting back to a system of equations, we find x = 400, y = , and z =.Inother words each meal should comprise 400 grams from food group A, 400 grams of food B, and 2000 grams of food C. (b) This time we re working with the system [ ] Replace Row 2 by Row 2 Row 1, then multiply Row 2 by 1. then divide both rows by. This gives [ ] Now replace Row 1 by Row 1 Row 2, and then divide each row by. The net result is Using z as the parameter we have (x, y, z) =( 800 z, 400,z). Page 5 of 6 A. Sontag February 1, 2004

6 Because none of the variables x, y, z can be negative, we must have 0 z 800. In other words, the number of grams of food group B must be 400, the number of grams of food group C can be any number between 0 and , and then the number of grams of food group A should be minus the number of grams of C. Note that the text says we can use any positive amount of food group C, but that s not right, since using too much C would cause the amount of A to become negative, which is impossible. (c) We could set up a new system of equations, using the original equations plus one new one expressing the cost condition. Since we already solved the original system, however, it s easier to reason as follows. The first three conditions force us to use 400 grams of A, 400 grams of B, and 2000 grams of C. With the given per-gram cost information, one such meal would cost, in cents, 2( )+(2000 ), which works out to more than 8 dollars. So, no, the stipulated cost is not possible with the given nutritional requirements. Page 6 of 6 A. Sontag February 1, 2004