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# MATH 1313, Exam 2 Study Guide Math 1313

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This 24 page Study Guide was uploaded by Theresa Nguyen on Sunday October 9, 2016. The Study Guide belongs to Math 1313 at University of Houston taught by moses sosa in Fall 2016. Since its upload, it has received 4 views. For similar materials see Finite Math with applications in Mathmatics at University of Houston.

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Review Test 2 Math 1313 Test 2 Review Sections 1.4, 1.5, Chapter 2 and Chapter 3 (0,150000) Example 1: Pull Company installed a new machine in one of its factories at a cost of $150,000. The machine is depreciated linearly over 10 years with no scrap value. Find an expression for the machine’s book value in the t-th year of use ( 0 < t < 10)(10,0) 2 −1 0−150000 m = 2 −1 = 10−0 = - 15000 y = mx + b V(t) = mt + Initial V(t) = -15000t + 150000 Example 2: A piece of equipment was purchased by a company for $10,000 and is assumed to have a scrap value of $3,000 in 5 years. If its value is depreciated linearly, find the value of the equipment after 3 years (0 < t < 5). (0,10000), (5,3000) y2−y1 3000−10000 m = x2−x1 = 5−0 =−1400 V(t) = mt + Initial V(t) = -1400t + 10000 V(3) = -1400(3) + 10000 = 5800 Example 3: A bicycle manufacturer experiences fixed monthly costs of $75,000 and fix costs of $75 per standard model bicycle produced. The bicycles sell for $125 each. a. What is the cost, revenue and profit functions? C(x) = 75x + 75000 R(x) = 125x P(x) = R(x) – C(x) = (S – P)x – F = (125 – 75)x – 75000 = 50x - 75000 b. What is the break-even point? (x,(B.E.Q., B.E.R.) R(x) = C(x) or P(x) = 0 R(1500) = 125(1500) 125x = 75x + 75000 = 187,500 50x = 75000 Break Even Revenue Break Even Quantity Break Even Point (1500,187500) Review Test 2 Example 4: Solve using Gauss-Jordon. 0 1 0 3 −R +R →R 0 1 0 3 −3R +R → R [1 −6 0−13] 1 3 3 [0 −9 −1−16] 2 1 1 0 1 0 1 −3 1 0 1 −3 0 −9 −1| 2]9R2+3 →3[0 0 −1|2]R 3R 3 −16 2 1 0 1−3 1 0 0 −1 0 1 0|2 −R3+1 →10 1 0|2 [0 0 1−2 0 0 1 −2 (-1,2,-2) Or X = -1, y = 2, z = -2 Example 5: Determine which of the following matrices are in row-reduced form. If a matrix is not in row-reduced form, state why. 1 0 −3 1 9 0 2 a. 0 1 −2 b. [1 0−3] c. 0 0 1 1 0 0|0 0 0 0|0 Yes No Yes Wrong Direction Example 6: The reduced form for the augmented matrix of a system with 3 exists.ns and 3 unknowns is given. Give the solution to the system, if it 1 0 −5 −3 1 0 0 0 1 0 1 1 a. [0 1 0 |1] b. [ ]1 0|1 c. 0 1 0 2] 0 0 0 6 0 0 1 3 → 0 = 6 x = 0 x + z = 1 x = 1 – z False y = 1 y = 2 z = ??? No Solution z = 3 Infinitely Many Unique Solution x = 1 – z, y = 2, z is any real number Review Test 2 Example 7: Find the value for x and y: 1 2 y−1 2 −4 −2 3 4 −3 1 2 =2 0 −1 x −1 ][ ]4 −3 4 4 1 * X – 3(4) = 2(4) 1 * (1) – 3(y – 1) = 2(- 4) X – 2 = 8 1 – 3y + 3 = -8 X = 20 -3y + 4 = -8 -3y = -12y = 4 Example 8: Given the foll(2 x 3)atrices (3 x 2)e product. 0 −2 1 1 −2 4 −1 0][ ] 1 −2 −1 (2 x 3)(3 x 2) 2 x 2 R1∗C1 R1∗C2 0 + 0 – 2 0 – 2 – 1 R2∗C1 R2∗C2 4 + 0 + 0 -8 – 1 + 0 −2 −3 = [4 −9] Example 9: Find the transpose of matrix A. 1 −4 3 A = [−2 7 4/3 → Rows Columns 1 −2 AT = −4 7 [3] 4/3 Review Test 2 Example 10: Find the inverse of matrix A. −3 4 [1 −2] −1 1 d −b A = D −c a ] D = ad – bc = -3(-2) – 4(1) = 6 – 4 = 2 = [−1/2 −3/2 Example 11: Solve the system of equations by using the inverse of the X – y = -4t matrix. 5x + 6y = 2 A = 1 −1 B = ( ) [5] 6 2 A−1 X = * B Ax = B −1 1 d −b 1 6 1 −4 A = D[−c a] = 11 −5 1(2) D = 1(6) – (-1)(5) = 6 + 5 = 11 1 6 1 −4 1 −24 2 =1 −22 −2 X = 11 −5 1( 2 = 11 20 2] 11[22 = (2) X = -2 or (-2,2) Y = 2 Review Test 2 Example 12: A vineyard produces two special wines a white and a red. A bottle of the white wine requires 14 pounds of grapes and one hour of processing time. A bottle of red wine requires 25 pounds of grapes and 2 hours of processing time. The vineyard has on hand 2,198 pounds of grapes and can allot 160 hours of processing time to the production of these wines. A bottle of the white wine makes $11.00 profit, while a bottle of the red wines makes $20.00 profit. Set-up the linear programming problem so that profit can be maximized. X – white wine y = red wine Grapes (lbs) 14 25 ≤ 2198 Processing (time) 1 2 ≤ 160 Profit 11 20 Max Profit = 11x + 20y 14x + 25y ≤ 2198 ≤ X + y 160 X,y ≥ 0 13. Solve the linear programming problem. Max P(x) = 3x + 7y St: 2x+5y ≤ 20 x+y ≤ 7 ≥ x, y 0 1 Line 2 ndLine X-int: (10,0) x-int: (7,0) y-int: (0,4) y-int: (0,7) 2x + 5y ≤ 20 x + y ≤ 7 5y ≤ -2x + 20 y ≤ 7 – x Pts Max P = 3x + 7y ≤ Y -2/5x + 4 (0,4) 3(0) + 7(4) = 28 -2/5x + 4 = 7 – x (5,2) 3(5) + 7(2) = 29 3/5x = 3 (7,0) 3(7) + 7(0) = 21 X = 5 (0,0) 3(0) + 7(0) = 0 y = 2 Optimal Value is 29 occurs @ (5,2) Review Test 2 14. Solve the linear programming problem. Min C(x) = x + 6y St: 3x+4y ≥ 36 ≥ 2x+y 14 x, y ≥ 0 st nd 1 Line 2 Line x-int: (12,0) x-int: (7,0) y-int: (0,9) y-int: (0,14) ≥ ≥ 3x + 4y 36 2x + y 14 4y ≥ -3x + 36 y ≥ 14 – 2x Pts Min x + 6y Y ≥ -3/4x + 9 (0,14) 0 + 6(14) = 84 -3/4x + 9 = 14 – 2x (4,6) 4 + 6(6) = 40 5/4x = 5 (12,0) 12 + 6(0) = 12 X = 4 Y = 6 Optimal Value is 12 occurs @ (12,0) Math 1313 Test 2 Supplemental Review 1. It costs a recording company $180,000 to prepare a new CD – recording costs, case design costs, etc. There are other costs such as materials, marketing and royalties that amount to $6 per CD. Suppose the CD is sold to music stores for $12. a. Write the cost function. C(x) = 6x + 180,000 b. Write the revenue function. R(x) = 12x c. Write the profit function. P(x) = R(x) – C(x) = (S – C)x – F = (12 – 6)x - 180,000 P(x) = 6x - 180,000 d. Is there a profit or a loss if the company sells 28,000 CDs? P(28,000) = 6(28,000) - 180,000 = -12,000; Loss of $12,000 e. What is the profit or loss if the company sells 50,000 CDs? P(50,000) = 6(50,000) – 180,000 = 120,000; Profit of $120,000 f. What is the break-even quantity? R(x) = C(x) 6x = 180,000 12x = 6x + 180,000 x = 30,000 units g. What is the break-even revenue? R(30,000) = 12(30,000) = $360,000 h. What is the break-even point? (30000, 360000) 2. Gallery of Furniture purchased 10 new delivery vans, to be depreciated linearly. Each van cost $60,000 and was sold for scrap value of $4,000 after four years. a. Write an expression for the value of one van at time t, (40≤≤t). (0, 60000) (4, 4000) y2−y1 4,000−60,000 −56,000 m = x −x = 4−0 = 4 = -14,000 2 1 Slope is -14,000, so the rate of depreciation is $14,000 Y = mx + b V(t) = mt + Initial V(t) = -14,000t + 60,000 b. Find the value of the van at the end of 3 years. V(3) = -14,000(3) + 60,000 = $18,000 3. Write the augmented matrix corresponding to the given system of equations. 8x – 4y – 2z = -10 8 −4 −2 −10 7y – 1/2z = 0 0 7 −1/| 0 [2 0 −8 −5 ] 2x – 8z = -5 Math 1313 Test 2 Review 1 4. Write the system of equations corresponding to the given augmented matrix. −9 3 8/7−1 [5 8 0 |0] 5 3 −1 4 -9x + 3y + 8/7z = -1 5x + 8y = 0 5x + 3y – z = 4 5. Indicate whether the matrix is in row-reduced form. 1 −2 0 a. 0 0|0] Yes b. 1 1|−1 No 0 1 10] 1 0 1 9 c. 0 1 0 6 Yes 0 0 0 0 1 0 02 d. 0 −9 1|1 No 1 0 0 2 e. 0 0 0 0 No 0 0 1|−4 0 1−5 f.[0 0|0 Yes 1 0 −2 0|2 0 1 1 | −3 g. 0 0 0 1|4 Yes 0 0 0 0|0 Math 1313 Test 2 Review 2 1 0 3 9 h. 0 1 −1 8 Yes [0 0 0|0 6. Solve the system of linear equations using the Gauss-Jordan elimination method. a. 5x + 3y = 9 -25 +3y = -8 1 3/5 [−2 1−8 ]1/5 R1→ R[−2 1 −85] 2 R1+R2→R2 1 3/5 9/5 →R 1 3/59/5 [0 11/5−22/5 5/11 R2 2[0 1|−2] -3/5 R1+2 →R2 1 0 3 x = 3, y = -2 or (3,-2) [0 1−2 ] b. 2x + 3y – 6z = -11 X – 2y + 3z = 9 3x + y = 7 1 −2 −3−11 R ↔R 2 −3 −6 9 3 1 0| 7 ] [ 23 1 0 |71] 1 −2 3 9 -2 R1+R2→R2 ; -3R1+R3→ 3 0 7 −12 −29 1/7 R2 →R 2 0 7 −9−20 ] 1 −2 3 9 1 0 −3/7 5/7 0 1 −12−29 R2+1 →R1 R2+3 →R30 1 −12/7 −29/7 [ 7 |7 ] 2 ; -7 [0 0 3 | 9 ] 0 7 −9 −20 Math 1313 Test 2 Review 3 1 0 −3/7 5/7 1/3 R3→ 3 0 1 −12/7−29/7 3/7 R3+1 → 1 ;12/3 2 +R2→R [0 0 1 | 3 ] 1 0 0 2 [ ]1 0|1 x = 2, y = 1, z = 3 or (2,1,3) 0 0 1 3 c. 2x + 3y = 2 X + 3y = -2 X – y = 3 2 3 2 1 3 −2 1 3 −2 1 3 |2 R1↔R 22 3|2 -2 R1+R2→R2 ; -R1+R3→ 30 −3| 6 [1 −1 3 1 −1 3 [0 −4 5 R 0 1 −2 R +R →R R +R →R 0 1 4 -1/3 [0 −4|52 -3 2 1 1 ; 4 2 3 3[0 0|3 No Solution d. 3y + 2z = 4 2x + 2y – z = 7 0 3 2 2 −1 −3 1 −1/2 −3/2 2 −1 −3 3 R1↔ 20 3 2 4 R1→ R10 3 2 342 [2 2 −1|7] [ 2 2 −1|7]1/2 [2 2 −1 |7 ] 1 −1/2 −3/2 3/2 1 −1/2 −3/2 3/2 - 2R1+R3→R30 3 2 4 1/3 R2 →R 20 1 2/3 4/3 0 3 2 |4 ] 0 3 2 | 4] 1 0 −7/6 13/6 1/2 R2+R1→R1 ; -3R2+R3→R3[ 1 2/3|4/3] 0 0 0 0 Infinitely many solutions X – 7/6z = 13/6→ x = 7/6z + 13/6 → Y + 2/3z = 4/3 y = -2/3z + 4/3 (7/6x + 13/6, -2/3z + 4/3, z), z is any real number Math 1313 Test 2 Review 4 7. The following augmented matrix in row-reduced form is equivalent to the augmented matrix of a certain system of linear equations. Use this result to solve the system of equations. 1 0 0 −3 a. [0 1 −1−17] Infinitely many solutions given by: X = -3 → Y – z = -17 y = -17 + z So, (-3, -17 + z, z) where z is any real number. 1 −1 3 b. [ ]3 |1 No Solution 0 0 −1/3 0 1 0 1|3 c. 0 0 1 −2 | [ ]0 0 0|0 0 0 0 0|0 Infinitely many solutions given by: → Y + w = 3 y = 3 – w Z – 2w = 4 → z = 4 + 2w So, (x, 3 – w, 4 + 2w, w) where x and w are any real numbers 1 0 0 1 d. 0 1 0−8 x = 1, y = -8, z = 7 or (1, -8, 7) [0 0 1 7 8. Solve for the variables in the matrix equation. 2 x−2 3 2 3 u 2 a. [ 2 4 y−2] [ ] 2 4 5 2 z −3 2 4 −3 2 2x – 2 = 3 3 = u y – 2 = 5 2z = 4 2x = 5 y = 7 z = 2 X = 5/2 Math 1313 Test 2 Review 5 x −2 + −2 z = 4 −2 b. [3][y][ −1 2 2u 4] x = 6 = 4 z = 0z = -2 y = 2 = 4 2 = 2u= 2u u = 1 1 2 y−1 2 −4 −u 1 2 3y−3 6 c. [ ] [ - 3 1 2 ] [ ] [ ] [1 = 3 4 - 3 6 ] = x −1 4 2 z+1 4 4 x −1 12 6 z+3 −4 −u [ ] −1 4 4 1 – (3y – 3) = -4 -3y + 4 = -4 x – 12 = 4 -1 – (6z + 3) = 4 z = -8/6 1 – 3y + 3 = -4 -3y = -8 x = 16 -1 – 6z – 3 = 4 z = -4/3 Y = 8/3 -4 – 6z = 4 -6z = 8 x+2 3z+1 5y 3x 2z 5y 10 −14 80 d. -2[4u 0 3] + 1/2 2u −5 6 ] = -1 [10 5/2 3] -2(x + 2) + ½(3x) = -1(10) -2(3x + 1) + ½(2z) = -1(-14) -2(5y) + ½(5y) = -1(80) -2x - 4 + 3/2x = -10 -6z – 2 + z = 14 -10y + 5/2y = -80 -1/2x - 4 = -10 -5z – 2 = 14 -15/2y = -80 -1/2x = -6 -5z = 16 y = -80 * -2/15 X = -6 * -2 z = -16/5 y = 32/3 X = 12 -2(4u) + ½(2u) = -1(10) -7u = -10 -8u + u = -10 u = 10/7 2 7 −8 8 −2 1 −8 10 17 1 0 11 −13 5 0 21 −7 −4 9. Let A =[ ] −4 −11 , B = [16 20 4] and. C = [ 9 −10 −14 ] 10 7 6 a. What is the size of A, B and C? The Matrix A is 4x3. The Matrix B is 3x3. The Matrix C is 3x3 b. Find a31? Findb22? Find c23? a31= 15, b22 = 5, c23 = -4 Math 1313 Test 2 Review 6 c. Identify the square matrix or matrices. Matrices B and C d. What is the transpose of A, B aTd C? 2 1 15 10 8 −13 16 −8 21 9 A = 7 0 −4 7 ;B ¿−2 5 20;C = 10 −7 −10 [−8 11 −11 6] [ 1 0 4] [17 −4 −14 ] e. Compute -11A + 3C, if possible. 2 7 −8 1 0 11 −8 10 17 -11 15 −4 −11 + 3 21 −7 −4 = Not possible [ 10 7 6 [9 −10 −14 ] f. Compute 8B – 3C 8 −2 1 −8 10 17 8(8) 8(−2) 8(1) 8 −13 5 0 - 3 21 −7 −4 = [(−13) 8(5) −8(0] - [16 20 4] [ 9 −10 −14] 8(16) 8(20) 8(4) 3(21) 3(−7) 3(−4) −104 −40 0 −63 −21 −12 [3(9) 3(−10) 3(−14) [ 128 160 32 [- 27 −30 −42] 64−(−24) −16−30 8−51 88 −46 −43 = −104−63 40−(−21) 0−(−12) = −167 61 12 [128−27 160−(−30) 32−(−42)[ 101 190 74] 10. Perform the indicated operations. 10 6 −6 0 −40 −24 −6 0 −3 4 −2 −1 12 −16 −2 −1 a. -4[8] 1 ] [ 10[12 = −32 −4 - 10 12 = −40−(−6) −24−0 12−(−2) −16−(−1) [−32−10 −4−12 ] −34 −24 = [−42 −16 Math 1313 Test 2 Review 7 5 7 −1 5 −2 −1 5 7 −1/2 5/2 b. 9 5 + 1/2 [−3 10 ] [ 9 −6 −15 ] [ ] [9 5 + −3/2 5 ] - −18 −9 [−54 −135 ] 5− −(−18) 7+ −(−9) 2 2 45/2 37/2 = [9− −(−54) 5+5−(−135) ] = 123/2 145] 2 11. Multiply, if possible. 2 3 −1 1 3 0 −2 −10 a. [4 8 3 2] 1 5 2 −3 0 1 [6] 7 −1(2+1(−2)+31)+0(6) ¿ −1 3)+1−10 +3(5+0(7) −4 2)+8−2 +3(1+2(6) −1 2 −4 3)+8−10 +3(5+2(7) 22)−3(−2)+01)+1(6) = [ ] −63 2( )−( −1) +( ) +1(7) ¿ 16 43 ¿ ¿ ¿ −2 3 2 3 −7 2 b. [2 4 −2][ ] 4 5 2 −3 2 4 −3 2 −2(3+3(2+2(4) −2 −7 +3(4+2(−3) −2 (2+3(5+2(2) [2 ( )4( )−2(4) 2(−7)+4( )−2(−3) 2( )+4( )−2(2] 2 ( )3( )+2(4) 2(−7)−3( )+2(−3) 2( )−3( )+2(2) 8 20 15 = [6 8 20] 8 −32 −7 −4 −11 10 9 −7 6 −4 11 c. [ 6 −7 3 15 −3 −6 −20 1 ] = Not Possible Math 1313 Test 2 Review 8 8 −9 11 −1 6 7 d. −4 3 −4 −7 ] [20 6 8(6−9(−4) 87)−9(−7) 84 119 116)−1(−4) 11(7−1(−7) 70 84 −4( )+3(−4) −4( )+3(−7)= −36 −49 [20( )+6(−4)20( )+6(−7) [96 98 −7 8 −9 4 6 5 4 7 e. 11 12 3 1 −5 −3 −7 −2 [ 5 16 −7 3 4 −7 −5 −10 ] −7(4+8(1−9(3) −7 6)+8−5)−9(4) −7(5+8 −3)−9(−7) −7(4+8(−7)−9(−5) −7 7+8 −2)−9(−10) [11( )+1( ) +3(3)( ) 6(+1) −5 +3(( )11(5 )12 −3 +3( )) (1 ) +12 −7 ( )−5( 1) 7 +12]−2 +3(−10) 5( )+1( ) −7(3)5( )+1( )5 −7(45( )+1( −) −7(−7)5( )+1( −) −7(−5)5( )+1( −) −7(−10) −47 −118 4 −39 25 = [15 −78 −26−−57 73] 12. Show that the two matrices are inverses of each other. a. 1 −3 and −2 3 [1 −2 [−1 1 −1 d −b a b A = −c a] A = [c d 1 −3 −2 3 1 0 −2 3 1 −3 1 0 [1 −2 −1 1[ ] = 0 1 and [−1 1 1 −2[ ] = 0 1 2 4 −2 1/2 −3 −4 −4 −6 1 −1/2 2 3 b. [ 3 5 −1 ] [nd −1 1 2 ] Math 1313 Test 2 Review 9 2 4 −2 1 0 0 1 2 −1 1/2 0 0 −4 −6 1 0 1 0 1/2R1→R 1 −4 −6 1 0 1 0 [3 5 −1 0 0 1 ] [3 5 −1 0 0 1 ] ½(2) = 1 ½(1) = ½ 4 R1+R2→R2 4(1) – 4 = 0 -3(1) + 3 = 0 R1+R3→ R3 ½(4) = 2 ½(0) = 0 -3 4(2) – 6 = 2 -3(2) + 5 = -1 ½(-2) = -1 ½(0) = 0 4(-1) + 1 = -3 -3(-1) – 1 = 2 0 = -3/2 4(1/2) + 0 = 2 -3(1/2) + 4(0) + 1 = 1 -3(0) + 0 = 0 4(0) + 0 = 0 -3(0) + 1 = 1 1 2 −1 1/2 0 0 1 2 −1 1/2 0 0 0 2 −3 2 1 0 1/2R →R 0 1 −3/2 1 1/2 0 [0 −1 2 −3/2 0 1 ] [ 2 20 −1 2 |−3/2 0 1] ½(0) = 0 -2(0) + 1 = 1 0 + 0 = 0 R2+R1→R 1 ½(2) = 1 -2 -2(1) + 2 = 0 1 – 1 = 0 ½(-3) = -3/2 R2+3 →R3 -2(-3/2) – 1 = 2 -3/2 + 2 = 1/2 ½(2) = 1 -2(1) + ½ = -3/21 – 3/2 = -1/2 ½(1) = ½ -2(1/2) + 0 = -1 ½ + 0 = 1/2 ½(0) = 0 -2(0) + 0 = 0 0 + 1 = 1 1 0 2 −3/2 −1 0 2R →R 1 0 2 −3/2 −1 0 [0 1 −3/2| 1 1/2 0] [3 30 1 −3/2| 1 1/2 0] 0 0 1/2 −1/2 1/2 1 0 0 1 −1 1 2 R +R → R 2(0) = 0 2(-1/2) = -1 -2 3 1 1 -2(0) + 1 = 1 3/2(0) + 0 = 0 2(0) = 0 2(1/2) = 1 3/2 R3+2 →R2 -2(0) + 0 = 0 3/2(0) + 1 = 1 2(1/2) = 1 2(1) = 2 -2(1) + 2 = 0 3/2(1) – 3/2 = 0 -2(-1) – 3/2 = ½ 3/2(-1) + 1 = -1/2 -2(1) – 1 = -3 3/2(1) + ½ = 2 Math 1313 Test 2 Review 10 -2(2) + 0 = -4 3/2(2) + 0 = 3 1 0 0 1/2 −3 −4 −1 1/2 −3 −4 [0 1 0|−1/22 3] A = [1/2 2 3] 0 0 1 −1 1 2 −1 1 2 2 4 −2 1/2 −3 −4 1/2 −3 −4 2 4 −2 −4 −6 1 −1/2 2 3 and −1/2 2 3 −4 −6 1 = [3 5 −1][−1 1 2 ] [ −1 1 2 [ 3 5 −1] 1 0 0 [ ]1 0 0 0 1 13. Find the inverse of the given matrix, if it exists. 2 −5 a. A =[1 −3 A−1 1/Dd −b a b = [−c a] A = c d D = ad – bc D = 2(-3) – (-5)(1) = -6 – (-5) = -6 + 5 = -1 −1 −3 5 −3 5 3 −5 A = 1/−−1 2 = -−1 2[ ]= 1 −2 10 2 b. B =[1] 5 B−1 1/Dd −b a b = [−c a] B = c d D = ad – bc D = 10(5) – 2(1) = 50 – 2 = 48 1/485 −2 5/48 −1/24 B = −1 10] [ −1/485/24] 3 −3 c. C =[2 −2 Math 1313 Test 2 Review 11 d −b a b C−1 = 1/D−c a ] C = [c d D = ad – bc D = 3(-2) – (-3)(2) = -6 – (-6) = -6 + 6 = 0 C−1 1/0−2 3 = [−2 3 = C does not have an inverse 2 −1 2 d. D = [−2 −2 1 1 −1 3 1 0 0 1 −1 3 1 0 0 2 1 2 0 1 0 R +R →R R +R → R0 3 −4 −2 1 0 [−2 −2 1 0 0 1 ] -2 1 2 2 ; 2 1 3 30 −4 7 |2 0 1] -2(1) + 2 = 0 -2(1) + 0 = -2 2(1) – 2 = 0 2(1) + 0 = 2 -2(-1) + 1 = 3 -2(0) + 1 = 1 2(-1) – 2 = -4 2(0) + 0 = 0 -2(3) + 2 = -4 -2(0) + 0 = 0 2(3) + 1 = 7 2(0) + 1 R →R 1/3 2 2 1/3(0) = 0 1/3(-4) = -4/3 1/3(1) = 1/3 1/3(3) = 1 1/3(-2) = -2/3 1/3(0) = 0 1 −1 3 1 0 0 0 1 −4/3 −2/3 1/3 0R2+R1→ 1 ; 4 R2+3 →R3 [0 −4 7 |2 0 1] 0 + 1 = 1 -2/3 + 1 = 1/3 4(0) + 0 = 0 4(- 2/3) + 2 = -2/3 1 – 14(1/3) + 0 = 4/30 = 1/3 4(1) – 4 = 0 -4/3 + 3 = 5/3 0 + 0 = 0 4(-4/3) + 7 = 5/3 4(0) + 1 = 1 1 0 5/3 1/3 1/3 0 1 0 5/3 1/3 1/3 0 0 1 −4/3 −2/3 1/3 0 R → R0 1 −4/3 −2/3 1/3 0 [0 0 5/3 −2/3 4/3 1] 3/5 3 30 0 1 −2/5 4/5 3/5] 3/5(0) = 0 3/5(5/3) = 1 3/5(4/3) = 4/5 −5/3R3+R1→R1 Math 1313 Test 2 Review 12 3/5(0) = 0 3/5(-2/3) = -2/5 3/5(1) = 3/5 4/3 R3+R2→R2 -5/3(0) + 1 = 1 -5/3(-2/5) + 1/3 = 1 4/3(0) + 0 = 0 4/3(- -5/3(0) + 0 = 0/5 -5/3(4/5) + 1/3 = -1 4/3(0) + 1 = 1 4/3(4/5) + 1/3 = 7/5 -5/3(1) + 5/3 = 0 -5/3(3/5) + 0 = -1 4/3(1) – 4/3 = 0 4/3(3/5) + 0 = 4/5 1 0 0 1 −1 −1 1 −1 −1 0 1 0 −6/5 7/5 4/5 D =¿ −6/5 7/54/5 [0 0 1 −2/5 4/5 3/5] −2/5 4/53/5] 14. Solve the system of linear equations using the inverse of the coefficient matrix. a.3x + 5y = 85 2 3 5 d −b A = [3 5 B = 8) A−1 = 1/D[−c a] D = ad – bc D = 2(5) – 3(3) = 10 – 9 = 1 −1 1/1 5 −3 5 −3 −1 5 −3 5 A = −3 2] [ −3 2 ] A * B = [−3 2](8 = 5( )−3(8) 1 (−3( )+2(8)= (1 X = 1, y = 1 b. 8x + 5y = 70 -x – 5y = 35 8 5 70 −1 1/D d −b A = −1 −5 ] B = 35) A = −c a] D = ad – D = 8(-5) – 5(-1) = -40 – (-5) = -40 + 5 = -35 −5 −5 5/35 5/35 1/7 1/7 A−1 = −1/35[1 8] [= −1/35 −8/35] [ −1/35 −8/35] Math 1313 Test 2 Review 13 1 1 1/7 1/7 ( )+ (35) A−1 * B = [ ]70 = 7 7 = ( ) x = 15, y −1/35 −8/35(35 (−1 70− 8 (35) −10 35 35 = -10 15. Use the method of corners to solve this LPP: Maximize Z = 5x + 7y subject to 6x + 3y 24 ≤ 3x + 6y 30 X ≥ 0 Y ≥ 0 ≤ 6x + 3y 24 x-int y-int 3y ≤ -6x + 24 6 x+0=24 0 + 3y = 24 Y ≤ -2x + 8 6x = 24 → x = 4 3y = 24 → y = 8 ≤ 3x + 6y 30 x-int y-int -2x + 8 = -1/2x + 5 6y ≤ -3x + 30 3x + 0 = 30 0 + 6y = 30 3 = 3/2x Y ≤ -1/2x + 5 3x = 30 → x = 10 6y = 30 → y = 5 x = 2 -2(2) + 8 = 4 y = 4 -1/2(2) + 5 = 4 Pts Max 5x + 7y A maximum of 38 occurs at (2,4) (0,5) 5(0) + 7(5) = 35 (2,4) 5(2) + 7(4) = 38 (4,0) 5(4) + 7(0) = 20 (0,0) 5(0) + 7(0) = 0 16. Use the method of corners to solve this LPP Minimize R = 200x + 35y subject to 4x + 3y 24 3x + 4y ≥ 8 X ≥ 0 Math 1313 Test 2 Review 14 Y ≥ 0 4x + 3y ≤ 24 x-int y-int 3y ≤ -4x + 24 4x + 0 = 24 0 + 3y = 24 Y ≤ -4/3x + 8 4x = 24 → x = 6 3y = 24 → y = 8 3x + 4y ≥ 8 x-int y-int ≥ 4y -3x + 8 3x + 0 = 8 0 + 4y = 8 ≥ → → Y -3/4x + 2 3x = 8 x = 8/3 = 2.7 4y = 8 y = 2 Pts Min 200x + 35y (0,8) 200(0) + 35(8) = 280 (0,2) 200(0) + 35(2) = 70 A minimum of 70 occurs at (0,2) (2.7,0) 200(2.7) + 35(0) = 540 (6,0) 200(6) + 35(0) = 1200 17. You can use two types of fertilizer in your orange grove, Best Food and Natural Nutri. Each bag of Best Food contains 8 pounds of nitrogen, 4 pounds of phosphoric acid, and 2 pounds of chlorine. Each bag of Natural Nutri contains 3 pounds of nitrogen, 4 pounds of phosphoric acid and 1 pound of chlorine. You know that the grove needs at least 1,000 pounds of phosphoric acid and at most 400 pounds of chlorine. If you want to minimize the amount of nitrogen added to the grove, how many bags of each type of fertilizer should be used? How much nitrogen will be added? Let x = # bags of Best Food; y = # bags of Natural Nutri X y Nitrogen 8 3 ≥ Phosphoric acid4 4 1,000 Chlorine 2 1 ≤ 400 The linear programming problem (LPP): Min N = 8x + 3y ≥ s.t. 4x + 4y 1,000 ≤ 2x + y 400 X ≥ 0 Y ≥ 0 st 1 Line 4x + 4y ≥ 1,000 x-int y-int Math 1313 Test 2 Review 15 4y ≥ -4x + 1,000 4x + 0 = 1,000 0 + 4y = 1,000 Y ≥ -x + 250 4x = 1,000 → x = 250 4y = 1,000 → y = 250 nd 2 Line 2x + y ≤ 400 x-int y-int Y ≤ -2x + 400 2x + 0 = 400 0 + y = 400 → 2x = 400 x = 200 y = 400 -x + 250 = -2x + 400 X = 150 y = 100 -150 + 250 = 100 -2(150) + 400 = 100 Pts Min N = 8x + 3y (0,400) 8(0) + 3(400) = 1200 (0,250) 8(0) + 3(250) = 750 Min = 750 occurs at (0,250) (150,100) 8(150) + 3(100) = 1500 You should use 0 bags of Best Food and 250 bags of Natural Nutri. The amount of nitrogen that will be added is 750 pounds. 18. Set up and solve the LPP: A manufacturing plant makes two types of inflatable boats, a two-person boat and a four-person boat. Each two-person boat requires 0.9 labor-hours from the cutting department and 0.8 labor hours from the assembly department. Each four-person boat requires 1.2 labor-hours from the cutting department and 1.8 labor-hours from the assembly department. The maximum labor-hours available per month for the cutting department and the assembly department are 780 and 950, respectively. The company makes a profit of $25 on each two-person boat and $50 on each four-person boat. How many of each type of boat should be manufactured in order to maximize profit? (Assume all boats manufactured are sold.) Let x = # of two-person boat; y = # of four-person boat X y ≤ Cutting 0.9 1.2 780 Assembly 0.8 1.8 ≤ 950 Profit 25x 50y Math 1313 Test 2 Review 16 The linear programming problem (LPP): Max P = 25x + 50y ≤ s.t. 0.9x + 1.2y 780 0.8x + 1.8y ≤ 950 X ≥ 0 Y ≥ 0 st 1 Line 0.9x + 1.2y ≤ 780 x-int y-int 1.2y ≤ -0.9x + 780 0.9x + 0 = 780 0 + 1.2y = 780 Y ≤ -0.75x + 650 0.9x = 780 → x = 866.7 1.2y = 780 → y = 650 nd 2 Line 0.8x + 1.8y ≤ 950 x-int y-int 1.8y ≤ -0.8x + 950 0.8x + 0 = 950 0 + 1.8y = 950 ≤ → Y -0.4x + 527.8 0.8x = 950 x = 1187.5 1.8y → = 950 y = 527.8 -0.75x + 650 = -0.4x + 527.8 122.2 = 0.35x X = 349.14 Y = 388.1 -0.75(349.14) + 650 = 388.1 -0.4(349.14) + 527.8 = 388.1 Pts Max P = 25x + 50y (0,527.8) 25(0) + 50(527.8) = 26390 (0,0) 25(0) + 50(0) = 0 Max = 28133.5 occurs at (866.7,0) 25(866.7) + 50(0) = 21667.5 (349.14,388.1) (349.14,388.1) 25(349.14) + 50(388.1) = 28133.5 They should produce 350 4-person boats and 388 2-person boats Math 1313 Test 2 Review 17

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