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 \n\u2234 Maximum value of Z is 16 at B (0, 4).<\/p>\nQuestion 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 \n\u2234 Maximum value of Z is – 12 at A (4, 0).<\/p>\n
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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 <\/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 \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
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Question 5. \nMaximize Z = 3x + 2y subject to x + 2y \u2264 10, 3x + y \u2264 15, x, y \u2265 0. \nSolution: \n \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 \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
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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 \nThe feasible region is shaded in the figure. We use the comer point method to find the maximum\/minimum value of Z. \n \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 \nThe feasible region is shaded in the figure. We use corner point method to find maxi-mum\/minimum value of Z. \n \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 \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 \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
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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 \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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n