{"id":2779,"date":"2023-03-03T14:24:01","date_gmt":"2023-03-03T13:24:01","guid":{"rendered":"http:\/\/matma.com.pl\/?p=2779"},"modified":"2023-05-06T10:04:36","modified_gmt":"2023-05-06T08:04:36","slug":"zadanie-31-0-6-zbior-rozszerzenie","status":"publish","type":"post","link":"http:\/\/matma.com.pl\/?p=2779","title":{"rendered":"zadanie 31 (0-6) zbi\u00f3r rozszerzenie"},"content":{"rendered":"<p>Obliczymy promie\u0144 podstawy oraz wysoko\u015b\u0107 sto\u017cka. W tym celu uzupe\u0142nimy nasz obrazek i skorzystamy z twierdzenia Pitagorasa<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 33.5341%;\"><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x^2+r^2=1^2\" alt=\"x^2+r^2=1^2\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x^2=1-r^2\" alt=\"x^2=1-r^2\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x=\\sqrt{1-r^2}\" alt=\"x=\\sqrt{1-r^2}\" align=\"absmiddle\" \/><\/p>\n<p>wysoko\u015b\u0107 sto\u017cka:<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x+1=\\sqrt{1-x^2}+1\" alt=\"x+1=\\sqrt{1-x^2}+1\" align=\"absmiddle\" \/><\/td>\n<td style=\"width: 66.4659%;\"><img fetchpriority=\"high\" decoding=\"async\" class=\" wp-image-2782 aligncenter\" src=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-03-100054-300x297.jpg\" alt=\"\" width=\"312\" height=\"309\" srcset=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-03-100054-300x297.jpg 300w, http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-03-100054-150x150.jpg 150w, http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-03-100054.jpg 482w\" sizes=\"(max-width: 312px) 100vw, 312px\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Wiemy, \u017ce d\u0142ugo\u015b\u0107 promienia podstawy sto\u017cka nie mo\u017ce by\u0107 liczb\u0105 ujemn\u0105, mamy zatem:<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 50%;\"><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r&gt;&amp;space;0\\;&amp;space;\\;&amp;space;\\wedge&amp;space;\\;&amp;space;\\;&amp;space;1-r^2&gt;&amp;space;0\" alt=\"r&gt; 0\\; \\; \\wedge \\; \\; 1-r^2&gt; 0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r&gt;&amp;space;0\\;&amp;space;\\;&amp;space;\\wedge&amp;space;\\;&amp;space;\\;&amp;space;\\left&amp;space;(1-r&amp;space;\\right&amp;space;)\\left&amp;space;(&amp;space;1+r&amp;space;\\right&amp;space;)&gt;&amp;space;0\" alt=\"r&gt; 0\\; \\; \\wedge \\; \\; \\left (1-r \\right )\\left ( 1+r \\right )&gt; 0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r\\in&amp;space;\\left&amp;space;(&amp;space;0,1&amp;space;\\right&amp;space;)\" alt=\"r\\in \\left ( 0,1 \\right )\" align=\"absmiddle\" \/><\/td>\n<td style=\"width: 50%;\"><img decoding=\"async\" class=\"size-medium wp-image-2787 aligncenter\" src=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-03-101053-300x156.jpg\" alt=\"\" width=\"300\" height=\"156\" srcset=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-03-101053-300x156.jpg 300w, http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-03-101053.jpg 473w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Wyznaczmy teraz funkcje opisuj\u0105c\u0105 obj\u0119to\u015b\u0107 sto\u017cka i jej dziedzin\u0119<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V(r)=\\frac{1}{3}\\pi&amp;space;r^2\\cdot&amp;space;\\left&amp;space;(&amp;space;1+\\sqrt{1-r^2}&amp;space;\\right&amp;space;)\" alt=\"V(r)=\\frac{1}{3}\\pi r^2\\cdot \\left ( 1+\\sqrt{1-r^2} \\right )\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?D_{V}:r\\in&amp;space;\\left&amp;space;(&amp;space;0,1&amp;space;\\right&amp;space;)\" alt=\"D_{V}:r\\in \\left ( 0,1 \\right )\" align=\"absmiddle\" \/><\/p>\n<p>Obliczymy nast\u0119pnie pochodn\u0105 i jej miejsca zerowe<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V'(r)=\\frac{1}{3}\\cdot&amp;space;2\\pi&amp;space;r\\cdot&amp;space;\\left&amp;space;(&amp;space;1+\\sqrt{1-r^2}&amp;space;\\right&amp;space;)+\\frac{1}{3}\\pi&amp;space;r^2\\cdot&amp;space;\\frac{1}{2\\sqrt{1-r^2}}\\cdot&amp;space;\\left&amp;space;(&amp;space;-2r&amp;space;\\right&amp;space;)\" alt=\"V'(r)=\\frac{1}{3}\\cdot 2\\pi r\\cdot \\left ( 1+\\sqrt{1-r^2} \\right )+\\frac{1}{3}\\pi r^2\\cdot \\frac{1}{2\\sqrt{1-r^2}}\\cdot \\left ( -2r \\right )\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V'(r)=\\frac{2\\pi&amp;space;r}{3}+\\frac{2\\pi&amp;space;r\\sqrt{1-r^2}}{3}\\&amp;space;-\\frac{\\pi&amp;space;r^3}{3\\sqrt{1-r^2}}\" alt=\"V'(r)=\\frac{2\\pi r}{3}+\\frac{2\\pi r\\sqrt{1-r^2}}{3}\\ -\\frac{\\pi r^3}{3\\sqrt{1-r^2}}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V'(r)=\\frac{2\\pi&amp;space;r\\sqrt{1-r^2}}{3\\sqrt{1-r^2}}+\\frac{2\\pi&amp;space;r\\sqrt{1-r^2}\\sqrt{1-r^2}}{3\\sqrt{1-r^2}}\\&amp;space;-\\frac{\\pi&amp;space;r^3}{3\\sqrt{1-r^2}}\" alt=\"V'(r)=\\frac{2\\pi r\\sqrt{1-r^2}}{3\\sqrt{1-r^2}}+\\frac{2\\pi r\\sqrt{1-r^2}\\sqrt{1-r^2}}{3\\sqrt{1-r^2}}\\ -\\frac{\\pi r^3}{3\\sqrt{1-r^2}}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V'(r&amp;space;)=\\frac{2\\pi&amp;space;r\\sqrt{1-r^2}+2\\pi&amp;space;r\\left&amp;space;|&amp;space;1-r^2\\right&amp;space;|-\\pi&amp;space;r^3}{3\\sqrt{1-r^2}}\" alt=\"V'(r )=\\frac{2\\pi r\\sqrt{1-r^2}+2\\pi r\\left | 1-r^2\\right |-\\pi r^3}{3\\sqrt{1-r^2}}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V'(r&amp;space;)=\\frac{2\\pi&amp;space;r\\sqrt{1-r^2}+2\\pi&amp;space;r&amp;space;\\left&amp;space;(1-r^2&amp;space;\\right&amp;space;)-\\pi&amp;space;r^3}{3\\sqrt{1-r^2}}\" alt=\"V'(r )=\\frac{2\\pi r\\sqrt{1-r^2}+2\\pi r \\left (1-r^2 \\right )-\\pi r^3}{3\\sqrt{1-r^2}}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V'(r&amp;space;)=\\frac{2\\pi&amp;space;r\\sqrt{1-r^2}+2\\pi&amp;space;r&amp;space;-2\\pi&amp;space;r^3-\\pi&amp;space;r^3}{3\\sqrt{1-r^2}}\" alt=\"V'(r )=\\frac{2\\pi r\\sqrt{1-r^2}+2\\pi r -2\\pi r^3-\\pi r^3}{3\\sqrt{1-r^2}}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V'(r&amp;space;)=\\frac{\\pi&amp;space;r\\left&amp;space;(2\\sqrt{1-r^2}+2&amp;space;-3&amp;space;r^2&amp;space;\\right&amp;space;)}{3\\sqrt{1-r^2}}\" alt=\"V'(r )=\\frac{\\pi r\\left (2\\sqrt{1-r^2}+2 -3 r^2 \\right )}{3\\sqrt{1-r^2}}\" align=\"absmiddle\" \/><\/p>\n<p>Obliczymy teraz miejsca zerowe pochodnej<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V'(r)=0\" alt=\"V'(r)=0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\frac{\\pi&amp;space;r\\left&amp;space;(2\\sqrt{1-r^2}+2&amp;space;-3r^2&amp;space;\\right&amp;space;)}{3\\sqrt{1-r^2}}=0\\;&amp;space;\\;&amp;space;\/\\cdot&amp;space;3\\left&amp;space;(&amp;space;\\sqrt{1-r^2}&amp;space;\\right&amp;space;)\" alt=\"\\frac{\\pi r\\left (2\\sqrt{1-r^2}+2 -3r^2 \\right )}{3\\sqrt{1-r^2}}=0\\; \\; \/\\cdot 3\\left ( \\sqrt{1-r^2} \\right )\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\pi&amp;space;r\\left&amp;space;(&amp;space;2\\sqrt{1-r^2}&amp;space;+2-3r^2\\right&amp;space;)\\left&amp;space;(&amp;space;1-r^2&amp;space;\\right&amp;space;)=0\" alt=\"\\pi r\\left ( 2\\sqrt{1-r^2} +2-3r^2\\right )\\left ( 1-r^2 \\right )=0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?2\\sqrt{1-r^2}&amp;space;+2-3r^2=0\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=0\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;\\left&amp;space;(&amp;space;1-r&amp;space;\\right&amp;space;)\\left&amp;space;(&amp;space;1+r&amp;space;\\right&amp;space;)=0\" alt=\"2\\sqrt{1-r^2} +2-3r^2=0\\; \\; \\vee \\; \\; r=0\\; \\; \\vee \\; \\; \\left ( 1-r \\right )\\left ( 1+r \\right )=0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?2\\sqrt{1-r^2}=3r^2-2\\;&amp;space;\\;&amp;space;\/^2\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=0\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=1\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=-1\" alt=\"2\\sqrt{1-r^2}=3r^2-2\\; \\; \/^2\\; \\; \\; \\; \\vee \\; \\; r=0\\; \\; \\vee \\; \\; r=1\\; \\; \\vee \\; \\; r=-1\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?4\\left&amp;space;(&amp;space;1-r^2&amp;space;\\right&amp;space;)=9r^4-12r^2+4\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=0\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=1\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=-1\" alt=\"4\\left ( 1-r^2 \\right )=9r^4-12r^2+4\\; \\; \\; \\; \\vee \\; \\; r=0\\; \\; \\vee \\; \\; r=1\\; \\; \\vee \\; \\; r=-1\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?4-4r^2=9r^4-12r^2+4\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=0\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=1\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=-1\" alt=\"4-4r^2=9r^4-12r^2+4\\; \\; \\; \\; \\vee \\; \\; r=0\\; \\; \\vee \\; \\; r=1\\; \\; \\vee \\; \\; r=-1\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?9r^4-8r^2=0\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=0\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=1\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=-1\" alt=\"9r^4-8r^2=0\\; \\; \\; \\; \\vee \\; \\; r=0\\; \\; \\vee \\; \\; r=1\\; \\; \\vee \\; \\; r=-1\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r^2\\left&amp;space;(&amp;space;9r^2-8&amp;space;\\right&amp;space;)=0\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=0\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=1\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=-1\" alt=\"r^2\\left ( 9r^2-8 \\right )=0\\; \\; \\; \\; \\vee \\; \\; r=0\\; \\; \\vee \\; \\; r=1\\; \\; \\vee \\; \\; r=-1\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r^2=0\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;9r^2=8\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=0\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=1\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=-1\" alt=\"r^2=0\\; \\; \\vee \\; \\; 9r^2=8\\; \\; \\; \\; \\vee \\; \\; r=0\\; \\; \\vee \\; \\; r=1\\; \\; \\vee \\; \\; r=-1\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r=0\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r^2=\\frac{8}{9}\\;&amp;space;\\;&amp;space;\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=0\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=1\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=-1\" alt=\"r=0\\; \\; \\vee \\; \\; r^2=\\frac{8}{9}\\; \\; \\; \\; \\vee \\; \\; r=0\\; \\; \\vee \\; \\; r=1\\; \\; \\vee \\; \\; r=-1\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r=0\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=\\frac{2\\sqrt{2}}{3}\\;&amp;space;\\;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=&amp;space;-\\frac{2\\sqrt{2}}{3}\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=0\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=1\\;&amp;space;\\;&amp;space;\\vee&amp;space;\\;&amp;space;\\;&amp;space;r=-1\" alt=\"r=0\\; \\; \\vee \\; \\; r=\\frac{2\\sqrt{2}}{3}\\; \\;\\vee \\; \\; r= -\\frac{2\\sqrt{2}}{3}\\; \\; \\vee \\; \\; r=0\\; \\; \\vee \\; \\; r=1\\; \\; \\vee \\; \\; r=-1\" align=\"absmiddle\" \/><\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 50%;\"><img decoding=\"async\" class=\"wp-image-2788 aligncenter\" src=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-03-120015-300x118.jpg\" alt=\"\" width=\"422\" height=\"166\" srcset=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-03-120015-300x118.jpg 300w, http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-03-120015.jpg 482w\" sizes=\"(max-width: 422px) 100vw, 422px\" \/><\/td>\n<td style=\"width: 50%;\"><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V'(r)&gt;0\\;&amp;space;\\;&amp;space;\\Leftrightarrow&amp;space;\\;&amp;space;\\;&amp;space;r\\in&amp;space;\\left&amp;space;(&amp;space;0,\\frac{2\\sqrt{2}}{3}&amp;space;\\right&amp;space;)\" alt=\"V'(r)&gt;0\\; \\; \\Leftrightarrow \\; \\; r\\in \\left ( 0,\\frac{2\\sqrt{2}}{3} \\right )\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V'(r)&lt;0\\;&amp;space;\\;&amp;space;\\Leftrightarrow&amp;space;\\;&amp;space;\\;&amp;space;r\\in&amp;space;\\left&amp;space;(&amp;space;\\frac{2\\sqrt{2}}{3},1&amp;space;\\right&amp;space;)\" alt=\"V'(r)&lt;0\\; \\; \\Leftrightarrow \\; \\; r\\in \\left ( \\frac{2\\sqrt{2}}{3},1 \\right )\" align=\"absmiddle\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Zatem funkcja <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V(r)\" alt=\"V(r)\" align=\"absmiddle\" \/> ro\u015bnie w przedziale <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r\\in&amp;space;\\left&amp;space;(&amp;space;0,\\frac{2\\sqrt{2}}{3}&amp;space;\\right&amp;space;]\" alt=\"r\\in \\left ( 0,\\frac{2\\sqrt{2}}{3} \\right ]\" align=\"absmiddle\" \/> oraz maleje w przedziale <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r\\in&amp;space;\\left&amp;space;[&amp;space;\\frac{2\\sqrt{2}}{3},1&amp;space;\\right&amp;space;)\" alt=\"r\\in \\left [ \\frac{2\\sqrt{2}}{3},1 \\right )\" align=\"absmiddle\" \/><\/p>\n<p>Oznacza to, \u017ce w punkcie <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?r=\\frac{2\\sqrt{2}}{3}\" alt=\"r=\\frac{2\\sqrt{2}}{3}\" align=\"absmiddle\" \/> funkcja osi\u0105ga maksimum.<\/p>\n<p>Obliczmy <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V\\left&amp;space;(&amp;space;\\frac{2\\sqrt{2}}{3}&amp;space;\\right&amp;space;)\" alt=\"V\\left ( \\frac{2\\sqrt{2}}{3} \\right )\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V\\left&amp;space;(&amp;space;\\frac{2\\sqrt{2}}{3}&amp;space;\\right&amp;space;)=\\frac{1}{3}\\pi&amp;space;\\cdot&amp;space;\\left&amp;space;(&amp;space;\\frac{2\\sqrt{2}}{3}&amp;space;\\right&amp;space;)^2\\cdot&amp;space;\\left&amp;space;(&amp;space;1+\\sqrt{1-\\left&amp;space;(&amp;space;\\frac{2\\sqrt{2}}{3}&amp;space;\\right&amp;space;)^2}&amp;space;\\right&amp;space;)&amp;space;=\\frac{1}{3}\\pi&amp;space;\\cdot&amp;space;\\frac{8}{9}\\cdot&amp;space;\\left&amp;space;(&amp;space;1+\\sqrt{1-\\frac{8}{9}}&amp;space;\\right&amp;space;)\" alt=\"V\\left ( \\frac{2\\sqrt{2}}{3} \\right )=\\frac{1}{3}\\pi \\cdot \\left ( \\frac{2\\sqrt{2}}{3} \\right )^2\\cdot \\left ( 1+\\sqrt{1-\\left ( \\frac{2\\sqrt{2}}{3} \\right )^2} \\right ) =\\frac{1}{3}\\pi \\cdot \\frac{8}{9}\\cdot \\left ( 1+\\sqrt{1-\\frac{8}{9}} \\right )\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?V\\left&amp;space;(&amp;space;\\frac{2\\sqrt{2}}{3}&amp;space;\\right&amp;space;)=\\frac{8}{27}\\pi&amp;space;\\cdot&amp;space;\\left&amp;space;(&amp;space;1+\\sqrt{\\frac{1}{9}}&amp;space;\\right&amp;space;)=\\frac{8}{27}\\pi&amp;space;\\cdot\\left&amp;space;(&amp;space;1+\\frac{1}{3}&amp;space;\\right&amp;space;)=\\frac{8}{27}\\pi&amp;space;\\cdot\\frac{4}{3}=\\frac{32}{81}\\pi\" alt=\"V\\left ( \\frac{2\\sqrt{2}}{3} \\right )=\\frac{8}{27}\\pi \\cdot \\left ( 1+\\sqrt{\\frac{1}{9}} \\right )=\\frac{8}{27}\\pi \\cdot\\left ( 1+\\frac{1}{3} \\right )=\\frac{8}{27}\\pi \\cdot\\frac{4}{3}=\\frac{32}{81}\\pi\" align=\"absmiddle\" \/><\/p>\n<p>Czyli najwi\u0119ksza obj\u0119to\u015b\u0107 sto\u017cka jest r\u00f3wna <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\frac{32}{81}\\pi\" alt=\"\\frac{32}{81}\\pi\" align=\"absmiddle\" \/> i otrzymujemy j\u0105 dla promienia d\u0142ugo\u015bci <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\frac{2\\sqrt{2}}{3}\" alt=\"\\frac{2\\sqrt{2}}{3}\" align=\"absmiddle\" \/><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Obliczymy promie\u0144 podstawy oraz wysoko\u015b\u0107 sto\u017cka. W tym celu uzupe\u0142nimy nasz obrazek i skorzystamy z twierdzenia Pitagorasa wysoko\u015b\u0107 sto\u017cka: Wiemy, \u017ce d\u0142ugo\u015b\u0107 promienia podstawy sto\u017cka nie mo\u017ce by\u0107 liczb\u0105 ujemn\u0105, mamy zatem: Wyznaczmy teraz funkcje opisuj\u0105c\u0105 obj\u0119to\u015b\u0107 sto\u017cka i jej dziedzin\u0119 Obliczymy nast\u0119pnie pochodn\u0105 i jej miejsca zerowe Obliczymy teraz miejsca zerowe pochodnej Zatem funkcja [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-2779","post","type-post","status-publish","format-standard","hentry","category-najnowsze-pliki"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.1 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>zadanie 31 (0-6) zbi\u00f3r rozszerzenie - Matma.com.pl<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"http:\/\/matma.com.pl\/?p=2779\" \/>\n<meta property=\"og:locale\" content=\"pl_PL\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"zadanie 31 (0-6) zbi\u00f3r rozszerzenie - Matma.com.pl\" \/>\n<meta property=\"og:description\" content=\"Obliczymy promie\u0144 podstawy oraz wysoko\u015b\u0107 sto\u017cka. W tym celu uzupe\u0142nimy nasz obrazek i skorzystamy z twierdzenia Pitagorasa wysoko\u015b\u0107 sto\u017cka: Wiemy, \u017ce d\u0142ugo\u015b\u0107 promienia podstawy sto\u017cka nie mo\u017ce by\u0107 liczb\u0105 ujemn\u0105, mamy zatem: Wyznaczmy teraz funkcje opisuj\u0105c\u0105 obj\u0119to\u015b\u0107 sto\u017cka i jej dziedzin\u0119 Obliczymy nast\u0119pnie pochodn\u0105 i jej miejsca zerowe Obliczymy teraz miejsca zerowe pochodnej Zatem funkcja [&hellip;]\" \/>\n<meta property=\"og:url\" content=\"http:\/\/matma.com.pl\/?p=2779\" \/>\n<meta property=\"og:site_name\" content=\"Matma.com.pl\" \/>\n<meta property=\"article:published_time\" content=\"2023-03-03T13:24:01+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-05-06T08:04:36+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/latex.codecogs.com\/gif.latex?x2+r2=12\" \/>\n<meta name=\"author\" content=\"lucynabartczak99\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Napisane przez\" \/>\n\t<meta name=\"twitter:data1\" content=\"lucynabartczak99\" \/>\n\t<meta name=\"twitter:label2\" content=\"Szacowany czas czytania\" \/>\n\t<meta name=\"twitter:data2\" content=\"9 minut\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"http:\/\/matma.com.pl\/?p=2779#article\",\"isPartOf\":{\"@id\":\"http:\/\/matma.com.pl\/?p=2779\"},\"author\":{\"name\":\"lucynabartczak99\",\"@id\":\"https:\/\/matma.com.pl\/#\/schema\/person\/b54229c3e035dea49ef8c166c5c63f1e\"},\"headline\":\"zadanie 31 (0-6) zbi\u00f3r rozszerzenie\",\"datePublished\":\"2023-03-03T13:24:01+00:00\",\"dateModified\":\"2023-05-06T08:04:36+00:00\",\"mainEntityOfPage\":{\"@id\":\"http:\/\/matma.com.pl\/?p=2779\"},\"wordCount\":106,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/matma.com.pl\/#organization\"},\"articleSection\":[\"najnowsze pliki\"],\"inLanguage\":\"pl-PL\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"http:\/\/matma.com.pl\/?p=2779#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"http:\/\/matma.com.pl\/?p=2779\",\"url\":\"http:\/\/matma.com.pl\/?p=2779\",\"name\":\"zadanie 31 (0-6) zbi\u00f3r rozszerzenie - 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