{"id":133901,"date":"2023-03-11T18:28:29","date_gmt":"2023-03-11T12:58:29","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=133901"},"modified":"2023-03-11T18:28:29","modified_gmt":"2023-03-11T12:58:29","slug":"solve-the-following-linear-programming-problem-graphically-minimise-z-60x-80y-subject-to-constraints","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/quiz\/solve-the-following-linear-programming-problem-graphically-minimise-z-60x-80y-subject-to-constraints","title":{"rendered":"Solve the following Linear Programming Problem graphically : Minimise: Z= 60x + 80y subject to constraints:"},"content":{"rendered":"<h2>Question: Solve the following Linear Programming Problem graphically : Minimise: Z= 60x + 80y subject to constraints:<br \/>\n3r + 4y \u2265 8<br \/>\n5r + 2y \u2265 11<br \/>\nx, y 20<\/h2>\n<p>&nbsp;<\/p>\n<h3>The correct answer is &#8211;<\/h3>\n<h4>To solve the linear programming problem graphically, we need to plot the feasible region and find the corner points of the region to determine the minimum value of the objective function Z = 60x + 80y.<\/h4>\n<h4>First, we plot the two constraint equations on a graph:<\/h4>\n<h4>3x + 4y = 8 (equation 1)<\/h4>\n<h4>5x + 2y = 11 (equation 2)<\/h4>\n<h4>To plot each line, we find their x and y intercepts.<\/h4>\n<ul>\n<li>\n<h4>For equation 1, when x = 0, y = 2, and when y = 0, x = 8\/3.<\/h4>\n<\/li>\n<li>\n<h4>For equation 2, when x = 0, y = 11\/2, and when y = 0, x = 11\/5.<\/h4>\n<\/li>\n<\/ul>\n<h4>We plot these intercepts and draw the lines passing through them. The feasible region is the area that satisfies the constraints and is bounded by these lines.<\/h4>\n<p>&nbsp;<\/p>\n<p>|<br \/>\n20 +&#8212;&#8212;&#8211;\/&#8212;&#8212;&#8211;+<br \/>\n| \/ | |<br \/>\n| \/ | |<br \/>\n15 +&#8212;&#8211;\/&#8212;+&#8212;&#8212;&#8211;+<br \/>\n| \/ | |<br \/>\n| \/ | |<br \/>\n10 +&#8211;\/&#8212;&#8212;+&#8212;&#8212;&#8211;+<br \/>\n| \/ | |<br \/>\n|\/ | |<br \/>\n5 +&#8212;&#8212;&#8212;-+&#8212;&#8212;\/-<br \/>\n0 2.75 5<br \/>\nx-axis<\/p>\n<p>&nbsp;<\/p>\n<div class=\"group w-full text-gray-800 dark:text-gray-100 border-b border-black\/10 dark:border-gray-900\/50 bg-gray-50 dark:bg-[#444654]\">\n<div class=\"text-base gap-4 md:gap-6 md:max-w-2xl lg:max-w-2xl xl:max-w-3xl p-4 md:py-6 flex lg:px-0 m-auto\">\n<div class=\"relative flex w-[calc(100%-50px)] flex-col gap-1 md:gap-3 lg:w-[calc(100%-115px)]\">\n<div class=\"flex flex-grow flex-col gap-3\">\n<div class=\"min-h-[20px] flex flex-col items-start gap-4 whitespace-pre-wrap\">\n<div class=\"markdown prose w-full break-words dark:prose-invert light\">\n<p>Next, we need to find the corner points of the feasible region by solving the intersection points of the lines. The corner points are (0, 2), (1.4, 1.5), and (2.75, 0).<\/p>\n<p>We then substitute the coordinates of each corner point into the objective function Z = 60x + 80y to find the minimum value of Z.<\/p>\n<ul>\n<li>Z at (0, 2): Z = 60(0) + 80(2) = 160<\/li>\n<li>Z at (1.4, 1.5): Z = 60(1.4) + 80(1.5) = 210<\/li>\n<li>Z at (2.75, 0): Z = 60(2.75) + 80(0) = 165<\/li>\n<\/ul>\n<p>Therefore, the minimum value of the objective function Z is 160, which occurs at point (0, 2). Therefore, x = 0 and y = 2, which satisfies the constraints, gives the minimum value of Z.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"flex justify-between\">\n<div class=\"text-gray-400 flex self-end lg:self-center justify-center mt-2 gap-3 md:gap-4 lg:gap-1 lg:absolute lg:top-0 lg:translate-x-full lg:right-0 lg:mt-0 lg:pl-2 visible\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bg-black mb-4 rounded-md\">\n<div class=\"flex items-center relative text-gray-200 bg-gray-800 px-4 py-2 text-xs font-sans\"><\/div>\n<\/div>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Question: Solve the following Linear Programming Problem graphically : Minimise: Z= 60x + 80y subject to constraints: 3r + 4y \u2265 8 5r + 2y \u2265 11 x, y 20 &nbsp; The correct answer is &#8211; To solve the linear programming problem graphically, we need to plot the feasible region and find the corner points [&hellip;]<\/p>\n","protected":false},"author":21842,"featured_media":132233,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12119],"tags":[],"class_list":{"0":"post-133901","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-quiz"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/133901","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21842"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=133901"}],"version-history":[{"count":2,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/133901\/revisions"}],"predecessor-version":[{"id":133904,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/133901\/revisions\/133904"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/132233"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=133901"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=133901"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=133901"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}