{"id":21479,"date":"2020-01-27T12:31:26","date_gmt":"2020-01-27T07:01:26","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=21479"},"modified":"2021-07-02T15:26:59","modified_gmt":"2021-07-02T09:56:59","slug":"2nd-puc-maths-question-bank-chapter-12-ex-12-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-12-ex-12-1\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1"},"content":{"rendered":"

Students can Download Maths Chapter 12 Linear Programming Ex 12.1 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1<\/h2>\n

2nd PUC Maths Linear Programming NCERT Text Book Questions and Answers Ex 12.1<\/h3>\n

Solve the following Linear Programming Problems graphically:<\/p>\n

Question 1.
\nMaximize Z = 3x + 4y subject to the constraints : x + y \u2264 4, x \u2265 0, y \u22650.
\nAnswer:
\nx + y = 4<\/p>\n\n\n\n\n
X<\/td>\n2<\/td>\n3<\/td>\n1<\/td>\n<\/tr>\n
y<\/td>\n2<\/td>\n1<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

put x = 0 y = 0
\nsince 0 + 0 \u2264 0, equation is true.
\n\u2234 This will be shaded towards 0.
\nAt A (4,0), Z = 3 x 4 = 12
\nAt B (0,4), Z = 4 x 4= 16
\n\u2234\u00a0 Z is maximised at B (0,4) = 16
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nMinimize Z = -3x + 4y
\nsubject to x + 2y \u2264 8; 3x + 2y \u2264 12; x \u2265 0, y \u2265 0.
\nAnswer:
\nx + 2y = 8<\/p>\n\n\n\n\n
X<\/td>\n2<\/td>\n4<\/td>\n0<\/td>\n<\/tr>\n
Y<\/td>\n3<\/td>\n2<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0 < 8
\ntrue towards 0<\/p>\n

3x +\u00a0 2y = 12<\/p>\n\n\n\n\n
X<\/td>\n2<\/td>\n4<\/td>\n0<\/td>\n<\/tr>\n
y<\/td>\n9<\/td>\n0<\/td>\n6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0 <12
\ntrue shaded towards 0
\nABCD is the solution region
\nx + 2y = 8
\nPut x = 0 \u21d2 A(-0,4)
\nPut y = 0 \u21d2 B(8,0)
\nAt A (0,4) \u21d2 Z= 16
\nAt B (2, 3) \u21d2 Z =-3 x 2+ 4 x 3 = 6
\nAt C (4, 0) \u21d2 Z = -12
\nAt D (0,0) \u21d2 z = 0
\nHence Z is minimum at C (4, 0) = -12
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nMaximise Z = 5x + 3y
\nsubject to 3x + 5y \u2264 15,5x +2y \u2264 10, x \u2265 0,y \u2265 0
\nAnswer:
\n3x + 5y \u2264 15<\/p>\n\n\n\n\n
X<\/td>\n0<\/td>\n5<\/td>\n<\/tr>\n
y<\/td>\n3<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0 <15
\ntrue, toward 0
\n\"2nd<\/p>\n

5x +2y = 10<\/p>\n\n\n\n\n
X<\/td>\n2<\/td>\n0<\/td>\n<\/tr>\n
y<\/td>\n0<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

ABCD is the solution region
\nAt A(0,3) ,Z = 3 x 3 = 9<\/p>\n

Question 4.
\nMinimum Z = 3x + 5y
\nsuch that x +3y \u2264 3, x+y \u2265 2, x, y\u00a0 \u22650.
\nAnswer:
\nx + 3y = 3<\/p>\n\n\n\n\n
X<\/td>\n3<\/td>\n0<\/td>\n<\/tr>\n
y<\/td>\n0<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0,3 false
\n\u2234 shading away from 0
\n\"2nd
\nx + y = 2<\/p>\n\n\n\n\n
X<\/td>\n0<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n
y<\/td>\n2<\/td>\n0<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0 ,2 false
\nshading away from 0
\nAway from 0
\ncorner points of feasible region are
\nA (0,2); Z = 5 x 2 = 10
\nB (1.5, 0.5); Z = 7
\nC (3,0); Z = 3 x 3 = 9
\n\u2234 Z is minimum at B (1.5,0.5) = 7.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nMaximise Z = 3x +2y
\nsubject to x + 2y \u2264 10, 3x + y \u2264 15, x, y \u2265 0.
\nAnswer:
\nx + 2y = 10<\/p>\n\n\n\n\n
X<\/td>\n0<\/td>\n6<\/td>\n4<\/td>\n<\/tr>\n
y<\/td>\n5<\/td>\n4<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0 < 10
\nTrue
\n\"2nd<\/p>\n

3y + y = 15<\/p>\n\n\n\n\n
X<\/td>\n5<\/td>\n4<\/td>\n3<\/td>\n<\/tr>\n
y<\/td>\n0<\/td>\n3<\/td>\n6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0 < 15
\nTrue<\/p>\n

ABCO is the solution region
\nAt A (0,5) Z = 2 x 5 = 10
\nAt B (4, 3) Z = 3 x 4+ 2 x 3 = 18
\nAt C (5,0) Z = 3 x 5 = 15
\n\u2234 Z is maximized at B (4, 3) = 18.<\/p>\n

Question 6.
\nMinimise Z = x + 2y
\nsubject to 2x + y \u2265 3, x + 2y \u2265 6, x, y \u2265 0
\nAnswer:
\n2x + y = 3<\/p>\n\n\n\n\n
X<\/td>\n1<\/td>\n0<\/td>\n<\/tr>\n
y<\/td>\n1<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0, 0 false
\n\"2nd<\/p>\n

x + 2y = 6<\/p>\n\n\n\n\n
X<\/td>\n6<\/td>\n0<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n0<\/td>\n3<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0 ,6 false
\nAn and above O is the solution region
\nAt A(0,3) Z = 2 x 3 = 6
\nAt B(6,0) Z = 6
\nZ is maximum at all points on line AB = 6
\nHere minimum of Z occurs at 2 points A & B.<\/p>\n

\"KSEEB<\/p>\n

Show that the minimum of Z occurs at more than two points.<\/p>\n

Question 7.
\nMinimise and Maximise Z = 5x + 10 y
\nsubject to x + 2y \u2264 120, x + y \u2265 60, x – 2y \u2265 0,x,y \u2265 0.
\nAnswer:
\nx + 2y = 120<\/p>\n\n\n\n\n
X<\/td>\n60<\/td>\n0<\/td>\n20<\/td>\n<\/tr>\n
Y<\/td>\n30<\/td>\n60<\/td>\n50<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0 ,120
\nTrue<\/p>\n

x + y = 60<\/p>\n\n\n\n\n
X<\/td>\n30<\/td>\n20<\/td>\n40<\/td>\n<\/tr>\n
Y<\/td>\n30<\/td>\n40<\/td>\n20<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0 >0
\nfalse
\n\"2nd<\/p>\n

x – 2y =0<\/p>\n\n\n\n\n
x<\/td>\n20<\/td>\n0<\/td>\n40<\/td>\n<\/tr>\n
y<\/td>\n10<\/td>\n0<\/td>\n20<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

40 – 20 > 0
\nABCD is the solution region
\nAt A (60, 30) Z = 5 x 60 + 10 x 30 = 600
\nAt B (40, 20) Z = 5 x 40+10 x 20 = 400
\nAt C (60,0) Z = 5 x 60 = 300
\nAt D (120,0) Z = 5 x 120 = 600
\nZ is minimized at C (60,0) = 300
\nZ is maximized at 2 points A (60,30), D (120,0)
\n\u2234 maximum value is 600 at all
\npoints on the line joining A (60,30) and D (120,0)<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nMinimise and Maximise Z = x + 2y
\nsubject to x + 2y \u2265 100, 2x – y \u2264 0, 2x + y \u2264 200; x, y \u2265 0.
\nAnswer:
\nx + 2y = 100<\/p>\n\n\n\n\n
x<\/td>\n100<\/td>\n50<\/td>\n20<\/td>\n<\/tr>\n
y<\/td>\n0<\/td>\n20<\/td>\n40<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0>100
\n0 – 0 < 0 False<\/p>\n

2x – y = 0<\/p>\n\n\n\n\n
x<\/td>\n100<\/td>\n20<\/td>\n10<\/td>\n<\/tr>\n
y<\/td>\n200<\/td>\n40<\/td>\n20<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 – 0 < 0
\nFalse<\/p>\n

2x +y = 200<\/p>\n\n\n\n\n
x<\/td>\n100<\/td>\n50<\/td>\n60<\/td>\n<\/tr>\n
y<\/td>\n0<\/td>\n100<\/td>\n80<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0 <200
\nTrue
\n\"2nd<\/p>\n

ABCD is the solution\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 .
\nAt A (0,200),\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Z = 2 x 200 = 400
\nAt B (50,100), Z = 50 + 2 x 100 = 250
\nAt C (20,40), Z = 20 + 2x 40= 100
\nAt D (0,50), Z = 2 x 50 =100
\nZ is maximised at A (0,200) = 400
\nZ is minimised at 2 points C (20,40) & D (0, 50) = 100<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nMaximise Z = -x + 2y, subject to the constraints:
\nx \u2265 3, x + y \u2265 5 , x + 2y \u2265 6, y \u2265 0
\nAnswer:
\nx + y = 5<\/p>\n\n\n\n\n
x<\/td>\n1<\/td>\n3<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n4<\/td>\n2<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0, 5
\nFalse<\/p>\n

0,3
\nFalse
\n\"2nd<\/p>\n

x + 2y = 6<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n4<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n3<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0
\nFalse<\/p>\n

ABC is the solution region
\nAt A (3,2)\u00a0 \u00a0 \u00a0 \u00a0Z =-3+2 x 2=1
\nAt B (4,1)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Z = -4 + 2 x 1 = -2
\nAt C (6, 0)\u00a0 \u00a0 \u00a0 Z = -6 = -6
\nZ is maximised at A (3,2) = 1 and
\nminimised at C (6,0) = -6<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nMaximise Z = x + y,
\nsubject to x – y \u2264 -1, -x +y \u2264 0, x, y \u2265 0.
\nAnswer:
\nx – y = -1<\/p>\n\n\n\n\n
x<\/td>\n1<\/td>\n0<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n2<\/td>\n1<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 – 0 ,-1
\nFalse<\/p>\n

x + 2y = 6<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n4<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n3<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

0 + 0 , 0 False,
\nSince there is no common shaded region, there is no feasible solution region. Hence maximum of Z does not exist.
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 12 Linear Programming Ex 12.1 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Maths Question Bank Chapter 12 Linear Programming Ex 12.1 2nd PUC Maths Linear Programming …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/21479"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=21479"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/21479\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=21479"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=21479"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=21479"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}