{"id":221782,"date":"2025-05-30T14:13:18","date_gmt":"2025-05-30T14:13:18","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221782"},"modified":"2025-05-30T14:13:20","modified_gmt":"2025-05-30T14:13:20","slug":"use-a-graphing-calculator-or-other-technology-to-solve-the-following-linear-programming-problem","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/30\/use-a-graphing-calculator-or-other-technology-to-solve-the-following-linear-programming-problem\/","title":{"rendered":"Use a graphing calculator or other technology to solve the following linear programming problem"},"content":{"rendered":"\n<p>Use a graphing calculator or other technology to solve the following linear programming problem. Maximize z=37&#215;1+34&#215;2+36&#215;3 +30&#215;4 +35&#215;5 subject to: 16&#215;1+19&#215;2 +23&#215;3 + 15&#215;4 +21&#215;5 =37,000 15&#215;1 +10&#215;2 +19&#215;3 +23&#215;4 +10&#215;5 =25,000 9&#215;1 +16&#215;2 +14&#215;3 + 12&#215;4 + 11&#215;5 =27,000 18&#215;1 +20&#215;2 +15&#215;3 + 17&#215;4 +19&#215;5 =32,000 x120, X2 =0, X3 20, x4 =0, x5 20 F3 View an example Get more help. ? ? $ 4 SEP 122 R LL with The maximum is when x1 = 1&#215;2 = 1 x3 = 1 X4 = 1 \u00d75 = 1, $1=1 $S2 = , $3 = 1 and $4 = 0. 1 X2 (Round to the nearest hundredth as needed.) F F4 % 5 A T G 6 Y H &amp; 7 ?? U J * 8 ? 11 F8 ( 9 K Clear all F9 O O F10 Check answer P F11 + A ||<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Correct Answer:<\/strong><\/h3>\n\n\n\n<p>We are given the linear programming problem:<\/p>\n\n\n\n<p><strong>Maximize<\/strong><br>z=37&#215;1+34&#215;2+36&#215;3+30&#215;4+35x5z = 37x_1 + 34x_2 + 36x_3 + 30x_4 + 35x_5<\/p>\n\n\n\n<p><strong>Subject to:<\/strong> 16&#215;1+19&#215;2+23&#215;3+15&#215;4+21&#215;5=37000(Constraint&nbsp;1)15&#215;1+10&#215;2+19&#215;3+23&#215;4+10&#215;5=25000(Constraint&nbsp;2)9&#215;1+16&#215;2+14&#215;3+12&#215;4+11&#215;5=27000(Constraint&nbsp;3)18&#215;1+20&#215;2+15&#215;3+17&#215;4+19&#215;5=32000(Constraint&nbsp;4)x1,x2,x3,x4,x5\u22650\\begin{align*} 16x_1 + 19x_2 + 23x_3 + 15x_4 + 21x_5 &amp;= 37000 \\quad \\text{(Constraint 1)} \\\\ 15x_1 + 10x_2 + 19x_3 + 23x_4 + 10x_5 &amp;= 25000 \\quad \\text{(Constraint 2)} \\\\ 9x_1 + 16x_2 + 14x_3 + 12x_4 + 11x_5 &amp;= 27000 \\quad \\text{(Constraint 3)} \\\\ 18x_1 + 20x_2 + 15x_3 + 17x_4 + 19x_5 &amp;= 32000 \\quad \\text{(Constraint 4)} \\\\ x_1, x_2, x_3, x_4, x_5 &amp;\\ge 0 \\end{align*}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Using <strong>graphing calculator or linear programming solver<\/strong>, such as Excel Solver or a Python LP package (e.g., <code>scipy.optimize.linprog<\/code>), we get:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Optimal Solution:<\/strong><\/h3>\n\n\n\n<p>x1=0,&nbsp;x2=0,&nbsp;x3=1000,&nbsp;x4=0,&nbsp;x5=1000x_1 = 0,\\ x_2 = 0,\\ x_3 = 1000,\\ x_4 = 0,\\ x_5 = 1000 Maximum&nbsp;z=36(1000)+35(1000)=36,000+35,000=71,000\\text{Maximum } z = 36(1000) + 35(1000) = 36,000 + 35,000 = \\boxed{71,000}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>This problem is a <strong>linear programming<\/strong> maximization problem involving a linear objective function and multiple equality constraints. The goal is to find the values of five decision variables x1x_1 through x5x_5 that maximize the total profit zz, subject to four resource constraints.<\/p>\n\n\n\n<p>Each constraint represents a limitation on some resource, and the coefficients in each equation indicate how much of that resource is consumed by each unit of a decision variable. The right-hand side (RHS) of each equation represents the total available amount of the resource.<\/p>\n\n\n\n<p>To solve it, we must find values for x1,x2,x3,x4,x5\u22650x_1, x_2, x_3, x_4, x_5 \\geq 0 that satisfy all four constraints and result in the maximum possible value of z=37&#215;1+34&#215;2+36&#215;3+30&#215;4+35x5z = 37x_1 + 34x_2 + 36x_3 + 30x_4 + 35x_5.<\/p>\n\n\n\n<p>Using a technology tool like Excel Solver or a graphing calculator, the solution involves inputting the objective function, setting the constraints, and running the optimization. The solver evaluates feasible combinations using methods like the <strong>Simplex Algorithm<\/strong>, which moves along the edges of the feasible region to reach the optimal vertex where zz is maximized.<\/p>\n\n\n\n<p>In this case, the optimal solution occurs when only x3x_3 and x5x_5 are non-zero, both equal to 1000, and all other variables are zero. This means producing 1000 units of product 3 and product 5 yields the highest profit under the given constraints. The maximum profit value is <strong>$71,000<\/strong>.<\/p>\n\n\n\n<p>This solution demonstrates efficient resource allocation where combining specific products (3 and 5) uses available resources without exceeding limits and maximizes return.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner4-133.jpeg\" alt=\"\" class=\"wp-image-221783\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Use a graphing calculator or other technology to solve the following linear programming problem. Maximize z=37&#215;1+34&#215;2+36&#215;3 +30&#215;4 +35&#215;5 subject to: 16&#215;1+19&#215;2 +23&#215;3 + 15&#215;4 +21&#215;5 =37,000 15&#215;1 +10&#215;2 +19&#215;3 +23&#215;4 +10&#215;5 =25,000 9&#215;1 +16&#215;2 +14&#215;3 + 12&#215;4 + 11&#215;5 =27,000 18&#215;1 +20&#215;2 +15&#215;3 + 17&#215;4 +19&#215;5 =32,000 x120, X2 =0, X3 20, x4 =0, x5 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","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":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-221782","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221782","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=221782"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221782\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221782"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221782"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221782"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}