{"id":32607,"date":"2022-03-30T10:00:03","date_gmt":"2022-03-30T04:30:03","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=32607"},"modified":"2022-03-30T10:29:24","modified_gmt":"2022-03-30T04:59:24","slug":"ncert-solutions-for-class-12-maths-chapter-12-ex-12-1","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-12-ex-12-1\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 12 Linear Programming Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-12-ex-12-1\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.1<\/h2>\n

\"NCERT<\/p>\n

Ex 12.1 Class 12 Question 1.<\/strong>
\nMaximize Z = 3x + 4y
\nsubject to the constraints:
\nx + y \u2264 4, x \u2265 0, y \u2265 0.
\nSolution:
\nThe objective function is Z = 3x + 4y
\nThe constraints are x + y \u2264 4, x \u2265 0, y \u2265 0.
\n\"NCERT
\n\u2234 Maximum value of Z is 16 at B (0, 4).<\/p>\n

Question 2.
\nMinimize Z = – 3x + 4y
\nsubject to x + 2y \u2264 8, 3x + 2y \u2264 12, x \u2265 0, y \u2265 0
\nSolution:
\nThe objective function is Z = – 3x + 4y The constraints are x + 2y \u2264 8, 3x + 2y \u2264 12, x \u2265 0, y \u2265 0
\n\"NCERT
\n\u2234 Maximum value of Z is – 12 at A (4, 0).<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nMaximize Z = 5x + 3y
\nsubject to 3x + 5y \u2264 15, 5x + 2y \u2264 10, x \u2265 0, y \u2265 0
\nSolution:
\nThe objective function is Z = 5x + 3y
\nThe constraints are 3x + 5y \u2264 15, 5x + 2y \u2264 10, x \u2265 0, y \u2265 0.
\nThe feasible region is shaded in the figure. The point B is obtained by solving the equation 5x + 2y = 10 and 3x + 5y = 15.
\n\"NCERT<\/p>\n

Question 4.
\nMinimize Z = 3x + 5y such that x + 3y \u2265 3, x + y \u2265 2, x, y \u2265 0.
\nSolution:
\nThe objective function is Z = 3x + 5y
\nThe constraints are x + 3y \u2265 3, x + y \u2265 2, x, y \u2265 0.
\nThe feasible region is shaded in the figure.
\n\"NCERT
\nFrom the table the minimum value of Z is 7. Since the feasible region is unbounded, 7 may or may not be the minimum value of Z. Consider the graph of the inequality 3x + 5y < 7. This half plane has no point in common with the feasible region. Hence the minimum value of Z is 7 at B (\\(\\frac { 3 }{ 2 }\\), \\(\\frac { 1 }{ 2 }\\))<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nMaximize Z = 3x + 2y subject to x + 2y \u2264 10, 3x + y \u2264 15, x, y \u2265 0.
\nSolution:
\n\"NCERT
\nThe objective function is Z = 3x + 5y
\nThe constraints are x + 2y < 10,
\n3x + y \u2264 15, x, y \u2265 0.
\nThe feasible region is shaded in the figure. We use the corner point method to find the maximum of Z
\nMaximum value of Z is 18 at B(4, 3)<\/p>\n

Question 6.
\nMinimize Z = x + 2y subject to 2x + y \u2265 3, x + 2y \u2265 6, x, y \u2265 0.
\nSolution:
\nConsider 2x + y \u2265 3
\nLet 2x + y = 3
\n\u21d2 y = 3 – 2x
\n\"NCERT
\n(0,0) is not contained in the required half plane as (0, 0) does not satisfy the in equation 2x + y \u2265 3.
\nAgain consider x + 2y \u2265 6
\nLet x + 2y = 6
\n\u21d2 \\(\\frac { x }{ 6 } +\\frac { y }{ 3 }\\) = 1
\nHere also (0,0) does not contain the required half plane. The double-shaded region XABY’ is the solution set. Its comers are A (6,0) and B (0,3). At A, Z = 6 + 0 = 6
\nAt B, Z = 0 + 2 \u00d7 3 = 6
\nWe see that at both points the value of Z = 6 which is minimum. In fact at every point on the line AB makes Z = 6 which is also minimum.<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nMinimise and Maximise Z = 5x + 10y subject to x + 2y \u2264 120, x + y \u2265 60, x – 2y \u2265 0, x, y \u2265 0
\nSolution:
\nThe objective function is Z = 5x + 10v.
\nThe constraints are x + 2y \u2264 120, x + y \u2265 60, x – 2y \u2265 0, x, y \u2265 0
\n\"NCERT
\nThe feasible region is shaded in the figure. We use the comer point method to find the maximum\/minimum value of Z.
\n\"NCERT
\nMinimum value of Z is 300 at A(60,0).
\nThe maximum value of Z is 600 at all points on the line joining B(120, 0) and C(60, 30).<\/p>\n

Question 8.
\nMinimize and maximize Z = x + 2y subject to x + 2y \u2265 100, 2x – y \u2264 0, 2x + y \u2264 200; x, y \u2265 0.
\nSolution:
\nThe objective function is Z = x + 2y
\nThe constraints are x + 2y \u2265 100, 2x – y \u2264 0, 2x + y \u2264 200; x, y \u2265 0
\n\"NCERT
\nThe feasible region is shaded in the figure. We use corner point method to find maxi-mum\/minimum value of Z.
\n\"NCERT
\nMaximum value of Z is 400 at B(0,200). The minimum value of Z is 100 at all points of the line joining the points (0,50) and (20,40).<\/p>\n

Question 9.
\nMaximize Z = – x + 2y, subject to the constraints: x \u2265 3, x + y \u2265 5, x + 2y \u2265 6, y \u2265 0
\nSolution:
\n\"NCERT
\nThe objective function is Z = – x + 2y
\nThe constraints are x \u2265 3, x + y \u2265 5, x + 2y \u2265 6, y \u2265 0
\nThe feasible region is shaded in the figure.
\n\"NCERT
\nFrom the table the maximum value of Z is 1. Since the feasible region is unbounded 1 may or may not be the maximum value of Z
\nConsider the graph of the inequality – x + 2y > 1. This half plane has points common with the feasible region. Hence there is no maximum value for Z.<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nMaximize Z = x + y subject to x – y \u2264 – 1, – x + y \u2264 0, x, y \u2265 0
\nSolution:
\nThe objective function is Z = x + y
\nThe constraints are x – y \u2264 – 1, – x + y \u2264 0, x, y \u2265 0
\n\"NCERT
\nThere is no point satisfying the constraints simultaneously. Thus the problem has no feasible region. Hence no maximum value for Z.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-12-ex-12-1\/ NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.1 Ex 12.1 Class 12 Question 1. Maximize Z = 3x + 4y subject to the constraints: x + …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-12-ex-12-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-12-ex-12-1\/ NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.1 Ex 12.1 Class 12 Question 1. 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