{"id":69083,"date":"2023-09-13T16:38:00","date_gmt":"2023-09-13T11:08:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=69083"},"modified":"2023-11-28T10:40:55","modified_gmt":"2023-11-28T05:10:55","slug":"rd-sharma-solutions-class-12-maths-chapter-18-exercise-18-5","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-18-exercise-18-5\/","title":{"rendered":"RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5 (Updated For 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-120861\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-18-Exercise-18.5.jpg\" alt=\"RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-18-Exercise-18.5.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Chapter-18-Exercise-18.5-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5<\/strong>: The RD Sharma book solution uses a structured approach to better understand and understand concepts. You can easily download the solution from the provided PDF. These solutions are designed by a group of experts in the field to eliminate students\u2019 doubts. The maximum and minimum values \u200b\u200bof functions in your domain are the main topics covered in this exercise. This exercise consists of two levels, in order of increasing difficulty.<\/p>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-12-solutions\/\" target=\"_blank\" rel=\"noopener\"><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Solutions for Class 12&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:16897,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;17&quot;:1}\">RD Sharma Solutions for Class 12 Maths All Chapters\u00a0<\/span><\/a>\u00a0<\/li>\n<\/ul>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69e77bc1533fa\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69e77bc1533fa\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-18-exercise-18-5\/#download-rd-sharma-solutions-class-12-maths-chapter-18-exercise-185-pdf\" title=\"Download RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5 PDF:\">Download RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5 PDF:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-18-exercise-18-5\/#access-rd-sharma-solutions-class-12-maths-chapter-18-exercise-185\" title=\"Access RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5\">Access RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-18-exercise-18-5\/#rd-sharma-solutions-class-12-maths-chapter-18-exercise-185-important-topics-from-the-chapter\" title=\"RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5: Important Topics From The Chapter\">RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5: Important Topics From The Chapter<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-18-exercise-18-5\/#faq-rd-sharma-solutions-class-12-maths-chapter-18-maxima-and-minima-exercise-185\" title=\"FAQ: RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5\">FAQ: RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-18-exercise-18-5\/#what-are-the-important-topics-included-in-rd-sharma-solutions-class-12-maths-chapter-18-maxima-and-minima-exercise-185\" title=\"What are the important topics included in RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5?\">What are the important topics included in RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-18-exercise-18-5\/#can-i-download-rd-sharma-solutions-class-12-maths-chapter-18-exercise-185-pdf-free\" title=\"Can I download RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5 PDF free?\">Can I download RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5 PDF free?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-18-exercise-18-5\/#what-are-the-benefits-of-studying-from-rd-sharma-solutions-class-12\" title=\"What are the benefits of studying from RD Sharma Solutions Class 12?\">What are the benefits of studying from RD Sharma Solutions Class 12?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-18-exercise-18-5\/#is-rd-sharma-enough-for-the-class-12-exams\" title=\"Is RD Sharma enough for the Class 12 Exams?\">Is RD Sharma enough for the Class 12 Exams?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-18-exercise-18-5\/#can-i-open-rd-sharma-solutions-class-12-maths-chapter-18-maxima-and-minima-exercise-185-pdf-on-my-smartphone\" title=\"Can I open RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5 PDF on my smartphone? \">Can I open RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5 PDF on my smartphone? <\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-12-maths-chapter-18-exercise-185-pdf\"><\/span>Download RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5 PDF:<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Maths-Chapter-18-Ex-18.5-1.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-12-Maths-Chapter-18-Ex-18.5-1.pdf\" data-wplink-edit=\"true\">RD-Sharma-Solutions-Class-12-Maths-Chapter-18-Ex-18.5<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-class-12-maths-chapter-18-exercise-185\"><\/span>Access RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>1. Determine two positive numbers whose sum is 15 and the sum of whose squares is minimum.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-14.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 14\" \/><\/p>\n<p>Which implies S is minimum when a = 15\/2 and b = 15\/2.<\/p>\n<p><strong>2. Divide 64 into two parts such that the sum of the cubes of two parts is minimum.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the two positive numbers be a and b.<\/p>\n<p>Given a + b = 64 \u2026 (1)<\/p>\n<p>We have, a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup>\u00a0is minima<\/p>\n<p>Assume, S = a<sup>3<\/sup>\u00a0+ b<sup>3<\/sup><\/p>\n<p>(From equation 1)<\/p>\n<p>S = a<sup>3<\/sup>\u00a0+ (64 \u2013 a)<sup>3<\/sup><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-15.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 15\" \/>of S<\/p>\n<p>Hence, the two number will be 32 and 32.<\/p>\n<p><strong>3. How should we choose two numbers, each greater than or equal to \u20132, whose sum is \u00bd so that the sum of the first and the cube of the second is minimum?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-16.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 16\" \/><\/p>\n<p>dS\/da = 0<\/p>\n<p>1 + 3(\u00bd \u2013 a)<sup>2<\/sup>\u00a0(-1) = 0<\/p>\n<p>1 \u2013 3(\u00bd \u2013 a)<sup>2<\/sup>\u00a0= 0<\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-17.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 17\" \/><\/p>\n<p><strong>4. Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the given two numbers be x and y. Then,<\/p>\n<p>x + y = 15 \u2026.. (1)<\/p>\n<p>y = (15 \u2013 x)<\/p>\n<p>Now we have z = x<sup>2<\/sup>\u00a0y<sup>3<\/sup><\/p>\n<p>z = x<sup>2<\/sup>\u00a0(15 \u2013 x)<sup>3<\/sup>\u00a0(from equation 1)<\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-18.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 18\" \/><\/p>\n<p><strong>5. Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm<sup>3<\/sup>, which has the minimum surface area?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let r and h be the radius and height of the cylinder, respectively. Then,<\/p>\n<p>Volume (V) of the cylinder = \u03c0r<sup>2<\/sup>\u00a0h<\/p>\n<p>\u27f9 100 = \u03c0r<sup>2<\/sup>\u00a0h<\/p>\n<p>\u27f9 h = 100\/ \u03c0r<sup>2<\/sup><\/p>\n<p>Surface area (S) of the cylinder = 2 \u03c0r<sup>2<\/sup>\u00a0+ 2 \u03c0r h = 2 \u03c0r<sup>2<\/sup>\u00a0+ 2 \u03c0r \u00d7 100\/ \u03c0r<sup>2<\/sup><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-19.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 19\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-20.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 20\" \/><\/p>\n<p><strong>6. A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by<\/strong><\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-21.gif\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 21\" \/><\/strong><\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-22.gif\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 22\" \/><\/strong><\/p>\n<p><strong>Find the point at which M is maximum in each case.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-23.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 23\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-24.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 24\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-25.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 25\" \/><\/p>\n<p><strong>7. A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the circle and the square is minimum?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Suppose the given wire, which is to be made into a square and a circle, is cut into two pieces of length x and y m respectively. Then,<\/p>\n<p>x + y = 28 \u21d2 y = (28 \u2013 x)<\/p>\n<p>We know that the perimeter of a square, 4 (side) = x<\/p>\n<p>Side = x\/4<\/p>\n<p>Area of square = (x\/4)<sup>2<\/sup>\u00a0= x<sup>2<\/sup>\/16<\/p>\n<p>Circumference of a circle, 2 \u03c0 r = y<\/p>\n<p>r = y\/ 2 \u03c0<\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-26.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 26\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-27.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 27\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-28.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 28\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-29.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 29\" \/><\/p>\n<p><strong>8.<\/strong>\u00a0<strong>A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the wire should be cut so that the sum of the areas of the square and triangle is minimum?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Suppose the wire, which is to be made into a square and a triangle, is cut into two pieces of length x and y respectively. Then,<\/p>\n<p>x + y = 20 \u21d2 y = (20 \u2013 x) \u2026\u2026 (1)<\/p>\n<p>We know that the perimeter of a square, 4 (side) = x<\/p>\n<p>Side = x\/4<\/p>\n<p>Area of square = (x\/4)<sup>2<\/sup>\u00a0= x<sup>2<\/sup>\/16<\/p>\n<p>Again we know that the perimeter of a triangle, 3 (side) = y.<\/p>\n<p>Side = y\/3<\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-30.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 30\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-31.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 31\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-32.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 32\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-33.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 33\" \/><\/p>\n<p>Hence, the wire of length 20 m should be cut into two pieces of lengths<\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-33a.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 33a\" \/><\/p>\n<p><strong>9. Given the sum of the perimeters of a square and a circle, show that the sum of their areas is least when one side of the square is equal to diameter of the circle.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us say the sum of the perimeter of the square and circumference of the circle be L<\/p>\n<p>Given the sum of the perimeters of a square and a circle.<\/p>\n<p>Assuming the side of the square = a and the radius of the circle = r<\/p>\n<p>Then, L = 4a + 2\u03c0r \u21d2 a = (L \u2013 2\u03c0r)\/4\u2026 (1)<\/p>\n<p>Let the sum of the area of the square and circle be S<\/p>\n<p>So, S = a<sup>2<\/sup>\u00a0+ \u03c0r<sup>2<\/sup><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-34.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 34\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-35.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 35\" \/><\/p>\n<p><strong>10. Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-36.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 36\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-37.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 37\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-38.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 38\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-39.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 39\" \/><\/p>\n<p><strong>11. Two sides of a triangle have lengths \u2018a\u2019 and \u2018b\u2019 and the angle between them is \u03b8. What value of \u03b8 will maximize the area of the triangle? Find the maximum area of the triangle also.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>It is given that two sides of a triangle have lengths a and b, and the angle between them is \u03b8.<\/p>\n<p>Let the area of the triangle be A<\/p>\n<p><img class=\"\" title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-41.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 41\" width=\"385\" height=\"207\" \/><br \/><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-40.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 40\" \/><\/p>\n<p><strong>12. A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Also, find this maximum volume<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-42.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 42\" \/><\/p>\n<p>Given side length of big square is 18 cm<\/p>\n<p>Let the side length of each small square be a.<\/p>\n<p>If by cutting a square from each corner and folding up the flaps we will get a cuboidal box with<\/p>\n<p>Length, L = 18 \u2013 2a<\/p>\n<p>Breadth, B = 18 \u2013 2a and<\/p>\n<p>Height, H = a<\/p>\n<p>Assuming, volume of box, V = LBH = a (18 \u2013 2a)<sup>2<\/sup><\/p>\n<p>Condition for maxima and minima is<\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-43.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 43\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-44.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 44\" \/><\/p>\n<p><strong>13. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off squares from each corners and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-45.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 45\" \/><\/p>\n<p>Given the length of the rectangle sheet = 45 cm<\/p>\n<p>Breath of rectangle sheet = 24 cm<\/p>\n<p>Let the side length of each small square be a.<\/p>\n<p>If by cutting a square from each corner and folding up the flaps we will get a cuboidal box with<\/p>\n<p>Length, L = 45 \u2013 2a<\/p>\n<p>Breadth, B = 24 \u2013 2a and<\/p>\n<p>Height, H = a<\/p>\n<p>Assuming, volume of box, V = LBH = (45 \u2013 2a)(24 \u2013 2a)(a)<\/p>\n<p>Condition for maxima and minima is<\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-46.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 46\" \/><\/p>\n<p>(45 \u2013 2a) (24 \u2013 2a) + (- 2) (24 \u2013 2a) (a) + (45 \u2013 2a) (- 2)(a) = 0<\/p>\n<p>4a<sup>2<\/sup>\u00a0\u2013 138a + 1080 + 4a<sup>2<\/sup>\u00a0\u2013 48a + 4a<sup>2<\/sup>\u00a0\u2013 90a = 0<\/p>\n<p>12a<sup>2<\/sup>\u00a0\u2013 276a + 1080= 0<\/p>\n<p>a<sup>2<\/sup>\u00a0\u2013 23a + 90= 0<\/p>\n<p>a = 5, 18<\/p>\n<p><img title=\"RD Sharma Solutions for Class 12 Maths Chapter 18\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/11\/rd-sharma-solutions-for-class-12-maths-chapter-18-maxima-and-minima-image-47.png\" alt=\"RD Sharma Solutions for Class 12 Maths Chapter 18 Maxima and Minima Image 47\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-12-maths-chapter-18-exercise-185-important-topics-from-the-chapter\"><\/span>RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5: Important Topics From The Chapter<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Some of the crucial topics of \u00a0<a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-18-maxima-and-minima\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 12 Maths Chapter 18<\/a> Exercise 18.5 are enlisted here.<\/p>\n<ul>\n<li>Maximum and minimum values of a function in its domain<\/li>\n<li>Definition and meaning of maximum<\/li>\n<li>Definition and meaning of minimum<\/li>\n<li>Local maxima<\/li>\n<li>Local minima<\/li>\n<li>First derivative test for local maxima and minima<\/li>\n<li>Higher-order derivative test<\/li>\n<li>Theorem and algorithm based on higher derivative test<\/li>\n<li>Point of inflection<\/li>\n<li>Point of inflection<\/li>\n<li>Properties of maxima and minima<\/li>\n<li>Maximum and minimum values in a closed interval<\/li>\n<li>Applied problems on maxima and minima<\/li>\n<\/ul>\n<p>We have included all the information regarding\u00a0<a href=\"http:\/\/cbseacademic.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5. If you have any query feel free to ask in the comment section.\u00a0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"faq-rd-sharma-solutions-class-12-maths-chapter-18-maxima-and-minima-exercise-185\"><\/span>FAQ: RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630079496300\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-important-topics-included-in-rd-sharma-solutions-class-12-maths-chapter-18-maxima-and-minima-exercise-185\"><\/span>What are the important topics included in RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can refer to the above article to know important topics included in RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630079508874\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-download-rd-sharma-solutions-class-12-maths-chapter-18-exercise-185-pdf-free\"><\/span>Can I download RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5 PDF free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can download RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630079525570\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-studying-from-rd-sharma-solutions-class-12\"><\/span>What are the benefits of studying from RD Sharma Solutions Class 12?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing these solutions, students can earn higher academic grades. Our experts solve these solutions with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630079526668\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-enough-for-the-class-12-exams\"><\/span>Is RD Sharma enough for the Class 12 Exams?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>RD Sharma has a variety of questions for class 12 board exams.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630080183920\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-open-rd-sharma-solutions-class-12-maths-chapter-18-maxima-and-minima-exercise-185-pdf-on-my-smartphone\"><\/span>Can I open RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5 PDF on my smartphone? <span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can open RD Sharma Solutions Class 12 Maths Chapter 18 Maxima and Minima Exercise 18.5 PDF on any device. <\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5: The RD Sharma book solution uses a structured approach to better understand and understand concepts. You can easily download the solution from the provided PDF. These solutions are designed by a group of experts in the field to eliminate students\u2019 doubts. The maximum and minimum &#8230; <a title=\"RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5 (Updated For 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-12-maths-chapter-18-exercise-18-5\/\" aria-label=\"More on RD Sharma Solutions Class 12 Maths Chapter 18 Exercise 18.5 (Updated For 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":120861,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2917,73716,73411,2985,73410],"tags":[3429,73223,73664],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/69083"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/244"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=69083"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/69083\/revisions"}],"predecessor-version":[{"id":513211,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/69083\/revisions\/513211"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/120861"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=69083"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=69083"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=69083"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}