{"id":3907,"date":"2022-04-05T18:26:28","date_gmt":"2022-04-05T18:26:28","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3907"},"modified":"2022-10-29T01:09:33","modified_gmt":"2022-10-29T01:09:33","slug":"functions-of-two-variables","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/functions-of-two-variables\/","title":{"raw":"Functions of Two Variables","rendered":"Functions of Two Variables"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Recognize a function of two variables and identify its domain and range.<\/li>\r\n \t<li>Sketch a graph of a function of two variables.<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe definition of a function of two variables is very similar to the definition for a function of one variable. The main difference is that, instead of mapping values of one variable to values of another variable, we map ordered pairs of variables to another variable.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nA <strong><span id=\"term149\" data-type=\"term\">function of two variables<\/span><\/strong><span id=\"term149\" data-type=\"term\"> [latex]z=f\\,(x,\\ y)[\/latex] maps each ordered pair [latex](x,\\ y)[\/latex] in a subset [latex]{\\bf{D}}[\/latex] of the real plane [latex]\\mathbb{R}^{2}[\/latex] to a unique real number [latex]z[\/latex]. The set [latex]{\\bf{D}}[\/latex] is called the <em>domain<\/em> of the function. The<em> range<\/em> of [latex]f[\/latex] is the set of all real numbers [latex]z[\/latex] that has at least one ordered pair [latex](x,\\ y)\\in{\\bf{D}}[\/latex] such that [latex]f\\,(x,\\ y)=z[\/latex] as shown in the following figure.<\/span>\r\n\r\n<\/div>\r\n[caption id=\"attachment_944\" align=\"aligncenter\" width=\"489\"]<img class=\"size-full wp-image-944\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/27163914\/4-1-1.jpeg\" alt=\"A bulbous shape is marked domain and it contains the point (x, y). From this point, there is an arrow marked f that points to a point z on a straight line marked range.\" width=\"489\" height=\"210\" \/> Figure 1.\u00a0The domain of a function of two variables consists of ordered pairs [latex](x,y)[\/latex].[\/caption]Determining the domain of a function of two variables involves taking into account any domain restrictions that may exist. Let's take a look.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Domains and Ranges for Functions of Two Variables<\/h3>\r\nFind the domain and range of each of the following functions:\r\n<ol>\r\n \t<li>[latex]f\\,(x,\\ y)=3x+5y+2[\/latex]<\/li>\r\n \t<li>[latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1167793361764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055154\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794055154\"]\r\n<ol>\r\n \t<li>This is an example of a linear function in two variables. There are no values or combinations of [latex]x[\/latex] and [latex]y[\/latex] that cause [latex]f\\,(x,\\ y)[\/latex] to be undefined, so the domain of [latex]f[\/latex] is [latex]\\mathbb{R}^{2}[\/latex]. To determine the range, first pick a value for [latex]z[\/latex]. We need to find a solution to the equation [latex]f\\,(x,\\ y)=z[\/latex], or [latex]3x-5y+2=z[\/latex]. One such solution can be obtained by first setting [latex]y=0[\/latex], which yields the equation [latex]3x+2=z[\/latex]. The solution to this equation is [latex]x=\\frac{z-2}{3}[\/latex], which gives the ordered pair [latex](\\frac{z-2}{3},\\ 0)[\/latex] as a solution to the equation [latex]f\\,(x,\\ y)=z[\/latex] for any value of [latex]z[\/latex]. Therefore, the range of the function is all real numbers, or [latex]\\mathbb{R}[\/latex].<\/li>\r\n \t<li>For the function [latex]g\\,(x,\\ y)[\/latex] to have a real value, the quantity under the square root must be nonnegative:\r\n<div style=\"text-align: center;\">[latex]9-x^{2}-y^{2}\\,\\geq\\,0.[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThis inequality can be written in the form\r\n<div style=\"text-align: center;\">[latex]x^{2}+y^{2}\\,\\leq\\,9.[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTherefore, the domain of [latex]g\\,(x,\\ y)[\/latex] is [latex]{\\{(x,\\ y)\\in\\mathbb{R}^{2}\\,|\\,x^{2}+y^{2}\\,\\leq\\,9\\}}[\/latex]. The graph of this set of points can be described as a disk of radius [latex]3[\/latex] centered at the origin. The domain includes the boundary circle as shown in the following graph.<\/li>\r\n<\/ol>\r\n<div>[caption id=\"attachment_946\" align=\"aligncenter\" width=\"417\"]<img class=\"size-full wp-image-946\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/27164019\/4-1-2.jpeg\" alt=\"A circle of radius three with center at the origin. The equation x2 + y2 = 9 is given.\" width=\"417\" height=\"422\" \/> Figure 2.\u00a0The domain of the function [latex]\\small{g(x,y)=\\sqrt{9-x^{2}-y^{2}}}[\/latex] is a closed disk of radius [latex]\\small{3}[\/latex].[\/caption]<\/div>\r\nTo determine the range of [latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}[\/latex] we start with a point [latex](x_0,\\ y_0)[\/latex] on the boundary of the domain, which is defined by the relation [latex]x^{2}+y^{2}=9[\/latex]. It follows that [latex]x^2_0+y^2_0=9[\/latex] and\r\n<div style=\"text-align: center;\">[latex]g\\,(x_0,\\ y_0)=\\sqrt{9-x^2_0-y^2_0}=\\sqrt{9-(x^2_0+y^2_0)}=\\sqrt{9-9}=0[\/latex].<\/div>\r\n&nbsp;\r\n\r\nIf [latex]x^2_0+y^2_0=0[\/latex] (in other words, [latex]x_0=y_0=0[\/latex]), then\r\n<div style=\"text-align: center;\">[latex]g\\,(x_0,\\ y_0)=\\sqrt{9-x^2_0-y^2_0}=\\sqrt{9-(x^2_0+y^2_0)}=\\sqrt{9-0}=3[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThis is the maximum value of the function. Given any value [latex]c[\/latex] between 0 and 3, we can find an entire set of points inside the domain of [latex]g[\/latex] such that [latex]g\\,(x,\\ y)=c[\/latex]:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc} \\hfill {\\sqrt{9-x^{2}-y^{2}}} &amp; =\\hfill &amp; {c} \\hfill \\\\ \\hfill {9-x^{2}-y^{2}} &amp; =\\hfill &amp; {c^{2}}\\hfill \\\\ \\hfill {x^{2}+y^{2}} &amp; =\\hfill &amp; {9-c^{2}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nSince [latex]9-c^{2}\\,&gt;\\,0[\/latex], this describes a circle of radius [latex]\\sqrt{9-c^{2}}[\/latex] centered at the origin. Any point on this circle satisfies the equation [latex]g\\,(x,\\ y)=c[\/latex]. Therefore, the range of this function can be written in interval notation as [latex][0,\\ 3][\/latex].\r\n<div><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the domain and range of the function [latex]f\\,(x,\\ y)=\\sqrt{36-9x^{2}-9y^{2}}[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1167793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793933114\"]\r\nThe domain is the shaded circle defined by the inequality [latex]9x^{2}+9y^{2}\\,\\leq\\,36[\/latex], which\u00a0has a circle of radius [latex]2[\/latex] as its boundary. The range is [latex][0,\\ 6][\/latex].\r\n\r\n[caption id=\"attachment_991\" align=\"alignnone\" width=\"342\"]<img class=\"size-full wp-image-991\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28231310\/4-1-tryitans1.jpeg\" alt=\"A circle of radius two with center at the origin. The equation x2 + y2 \u2264 4 is given.\" width=\"342\" height=\"347\" \/> Figure 3.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186147&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=UaWAuFLTy98&amp;video_target=tpm-plugin-8c5jetnr-UaWAuFLTy98\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.1_transcript.html\">transcript for \u201cCP 4.1\u201d here (opens in new window).<\/a><\/center>\r\n<h2>Graphing Functions of Two Variables<\/h2>\r\nSuppose we wish to graph the function [latex]z=(x,\\ y)[\/latex]. This function has two independent variables ([latex]x[\/latex] and [latex]y[\/latex]) and one dependent variable ([latex]z[\/latex]). When graphing a function [latex]y=f\\,(x)[\/latex] of one variable, we use the Cartesian plane. We are able to graph any ordered pair [latex](x,\\ y)[\/latex] in the plane, and every point in the plane has an ordered pair [latex](x,\\ y)[\/latex] associated with it. With a function of two variables, each ordered pair [latex](x,\\ y)[\/latex] in the domain of the function is mapped to a real number [latex]z[\/latex]. Therefore the graph of the function [latex]f[\/latex] consists of ordered pairs [latex](x,\\ y,\\ z)[\/latex]. The graph of a function [latex]z=f\\,(x,\\ y)[\/latex] of two variables is called a\u00a0<strong>surface<\/strong>.\r\n\r\nTo understand more completely the concept of plotting a set of ordered triples to obtain a surface in three-dimensional space, imagine the [latex]x,\\ y)[\/latex] coordinate system laying flat. Then, every point in the domain of the function [latex]f[\/latex] has a unique [latex]z[\/latex]-value associated with it. If [latex]z[\/latex] is positive, then the graphed point is located above the [latex]xy[\/latex]-plane. The set of all the graphed points becomes the two-dimensional surface that is the graph of the function [latex]f[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing Functions of Two Variables<\/h3>\r\nCreate a graph of each of the following functions:\r\n<ol>\r\n \t<li>[latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}[\/latex]<\/li>\r\n \t<li>[latex]f\\,(x,\\ y)=x^{2}+y^{2}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1267793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1267793933114\"]\r\n<ol>\r\n \t<li>In the previous example, we determined that the domain of [latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}[\/latex] is [latex]\\{(x,\\ y)\\in\\mathbb{R}^{2}\\,|\\,x^{2}+y^{2}\\,\\leq\\,9\\}[\/latex] and the range is [latex]{z\\in\\mathbb{R}^{2}\\,|\\,0\\,\\leq\\,z\\,\\leq\\,3}[\/latex]. When [latex]x^{2}+y^{2}=9[\/latex] we have [latex]g\\,(x,\\ y)=0[\/latex]. Therefore, any point on the circle of radius [latex]3[\/latex] centered at the origin in the [latex]xy[\/latex]-plane maps to [latex]z=0[\/latex] in [latex]\\mathbb{R}^3[\/latex]. If [latex]x^{2}+y^{2}=8[\/latex], then [latex]g\\,(x,\\ y)=1[\/latex], so any point on the circle of radius [latex]2\\sqrt{2}[\/latex] centered at the origin in the [latex]xy[\/latex]-plane maps to [latex]z=1[\/latex] in [latex]\\mathbb{R}^{3}[\/latex]. As [latex]x^{2}+y^{2}[\/latex] gets closer to zero, the value of [latex]z[\/latex] approaches [latex]3[\/latex]. When [latex]x^{2}+y^{2}=0[\/latex], then [latex]g\\,(x,\\ y)=3[\/latex]. This is the origin in the [latex]xy[\/latex]-plane. If [latex]x^{2}+y^{2}[\/latex] is equal to any other value between [latex]0[\/latex] and [latex]9[\/latex], then [latex]g\\,(x,\\ y)[\/latex] equals some other constant between [latex]0[\/latex] and [latex]3[\/latex]. The surface described by this function is a hemisphere centered at the origin with radius\u00a0[latex]3[\/latex] as shown in the following graph.\r\n<div>\r\n\r\n[caption id=\"attachment_949\" align=\"aligncenter\" width=\"597\"]<img class=\"size-full wp-image-949\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/27164347\/4-1-3.jpeg\" alt=\"A hemisphere with center at the origin. The equation z = g(x, y) = the square root of the quantity (9 \u2013 x2 \u2013 y2) is given.\" width=\"597\" height=\"612\" \/> Figure 4. <span class=\"os-caption\">Graph of the hemisphere represented by the given function of two variables.<\/span>[\/caption]\r\n\r\n<\/div><\/li>\r\n \t<li>This function also contains the expression [latex]x^{2}+y^{2}[\/latex]. Setting this expression equal to various values starting at zero, we obtain circles of increasing radius. The minimum value of [latex]f\\,(x,\\ y)=x^{2}+y^{2}[\/latex] is zero (attained when [latex]x=y=0[\/latex]). When [latex]x=0[\/latex], the function becomes [latex]z=y^{2}[\/latex], and when [latex]y=0[\/latex], then the function becomes [latex]z=x^{2}[\/latex]. These are cross-sections of the graph, and are parabolas. Recall from Introduction to Vectors in Space that the name of the graph of [latex]f\\,(x,\\ y)=x^{2}+y^{2}[\/latex] is a\u00a0<em>paraboloid<\/em>. The graph of [latex]f[\/latex] appears in the following graph.\r\n<div>\r\n\r\n[caption id=\"attachment_951\" align=\"aligncenter\" width=\"585\"]<img class=\"size-full wp-image-951\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/27164441\/4-1-4.jpeg\" alt=\"A paraboloid with vertex at the origin. The equation z = f(x, y) = x2 + y2 is given.\" width=\"585\" height=\"574\" \/> Figure 5.\u00a0A paraboloid is the graph of the given function of two variables.[\/caption]\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Nuts and Bolts<\/h3>\r\nA profit function for a hardware manufacturer is given by\r\n<p style=\"text-align: center;\">[latex]f\\,(x,\\ y)=16-(x-3)^{2}-(y-2)^{2}[\/latex],<\/p>\r\nwhere [latex]x[\/latex] is the number of nuts sold per month (measured in thousands) and [latex]y[\/latex] represents the number of bolts sold per month (measured in thousands). Profit is measured in thousands of dollars. Sketch a graph of this function.\r\n\r\n[reveal-answer q=\"fs-id1167794933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794933114\"]\r\nThis function is a polynomial function in two variables. The domain of [latex]f[\/latex] consists of [latex](x,\\ y)[\/latex] coordinate pairs that yield a nonnegative profit:\r\n<div style=\"text-align: center;\">[latex]16-(x-3)^{2}-(y-2)^{2}\\,\\geq\\,0[\/latex]<\/div>\r\n&nbsp;\r\n<div style=\"text-align: center;\">[latex](x-3)^{2}-(y-2)^{2}\\,\\leq\\,16.[\/latex]<\/div>\r\n&nbsp;\r\n<p style=\"text-align: left;\">This is a disk of radius\u00a0[latex]4[\/latex] centered at [latex](3,\\ 2)[\/latex]. A further restriction is that both [latex]x[\/latex] and [latex]y[\/latex] must be nonnegative. When [latex]x=3[\/latex] and [latex]y=2[\/latex], [latex]f\\,(x,\\ y)=16[\/latex]. Note that it is possible for either value to be a noninteger; for example, it is possible to sell\u00a0[latex]2.5[\/latex] thousand nut in a month. The domain, therefore, contains thousands of points, so we can consider all points within the disk. For any [latex]z&lt;16[\/latex], we can solve the equation [latex]f\\,(x,\\ y)=z[\/latex]:<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]16-(x-3)^{2}-(y-2)^{2}=z[\/latex]<\/div>\r\n&nbsp;\r\n<div style=\"text-align: center;\">[latex](x-3)^{2}-(y-2)^{2}=16-z[\/latex]<\/div>\r\n&nbsp;\r\n<p style=\"text-align: left;\">Since [latex]z&lt;16[\/latex], we know that [latex]16-z&gt;0[\/latex], so the previous equation describes a circle with radius [latex]\\sqrt{16-z}[\/latex] centered at the point [latex](3,\\ 2)[\/latex]. Therefore, the range of [latex]f\\,(x,\\ y)[\/latex] is [latex]\\{z\\in\\mathbb{R}\\,|z\\,\\leq\\,16\\}[\/latex]. The graph of [latex]f\\,(x,\\ y)[\/latex] is also a paraboloid, and this paraboloid points downward as shown.<\/p>\r\n\r\n\r\n[caption id=\"attachment_953\" align=\"aligncenter\" width=\"496\"]<img class=\"size-full wp-image-953\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/27164529\/4-1-5.jpeg\" alt=\"A paraboloid center seemingly on the positive z axis. The equation z = f(x, y) = 16 \u2013 (x \u2013 3)2 \u2013 (y \u2013 2)2 is given.\" width=\"496\" height=\"517\" \/> Figure 6.\u00a0The graph of the given function of two variables is also a paraboloid.[\/caption]\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Recognize a function of two variables and identify its domain and range.<\/li>\n<li>Sketch a graph of a function of two variables.<\/li>\n<\/ul>\n<\/div>\n<p>The definition of a function of two variables is very similar to the definition for a function of one variable. The main difference is that, instead of mapping values of one variable to values of another variable, we map ordered pairs of variables to another variable.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>A <strong><span id=\"term149\" data-type=\"term\">function of two variables<\/span><\/strong><span id=\"term149\" data-type=\"term\"> [latex]z=f\\,(x,\\ y)[\/latex] maps each ordered pair [latex](x,\\ y)[\/latex] in a subset [latex]{\\bf{D}}[\/latex] of the real plane [latex]\\mathbb{R}^{2}[\/latex] to a unique real number [latex]z[\/latex]. The set [latex]{\\bf{D}}[\/latex] is called the <em>domain<\/em> of the function. The<em> range<\/em> of [latex]f[\/latex] is the set of all real numbers [latex]z[\/latex] that has at least one ordered pair [latex](x,\\ y)\\in{\\bf{D}}[\/latex] such that [latex]f\\,(x,\\ y)=z[\/latex] as shown in the following figure.<\/span><\/p>\n<\/div>\n<div id=\"attachment_944\" style=\"width: 499px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-944\" class=\"size-full wp-image-944\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/27163914\/4-1-1.jpeg\" alt=\"A bulbous shape is marked domain and it contains the point (x, y). From this point, there is an arrow marked f that points to a point z on a straight line marked range.\" width=\"489\" height=\"210\" \/><\/p>\n<p id=\"caption-attachment-944\" class=\"wp-caption-text\">Figure 1.\u00a0The domain of a function of two variables consists of ordered pairs [latex](x,y)[\/latex].<\/p>\n<\/div>\n<p>Determining the domain of a function of two variables involves taking into account any domain restrictions that may exist. Let&#8217;s take a look.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Domains and Ranges for Functions of Two Variables<\/h3>\n<p>Find the domain and range of each of the following functions:<\/p>\n<ol>\n<li>[latex]f\\,(x,\\ y)=3x+5y+2[\/latex]<\/li>\n<li>[latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}[\/latex]<\/li>\n<\/ol>\n<div id=\"fs-id1167793361764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794055154\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794055154\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>This is an example of a linear function in two variables. There are no values or combinations of [latex]x[\/latex] and [latex]y[\/latex] that cause [latex]f\\,(x,\\ y)[\/latex] to be undefined, so the domain of [latex]f[\/latex] is [latex]\\mathbb{R}^{2}[\/latex]. To determine the range, first pick a value for [latex]z[\/latex]. We need to find a solution to the equation [latex]f\\,(x,\\ y)=z[\/latex], or [latex]3x-5y+2=z[\/latex]. One such solution can be obtained by first setting [latex]y=0[\/latex], which yields the equation [latex]3x+2=z[\/latex]. The solution to this equation is [latex]x=\\frac{z-2}{3}[\/latex], which gives the ordered pair [latex](\\frac{z-2}{3},\\ 0)[\/latex] as a solution to the equation [latex]f\\,(x,\\ y)=z[\/latex] for any value of [latex]z[\/latex]. Therefore, the range of the function is all real numbers, or [latex]\\mathbb{R}[\/latex].<\/li>\n<li>For the function [latex]g\\,(x,\\ y)[\/latex] to have a real value, the quantity under the square root must be nonnegative:\n<div style=\"text-align: center;\">[latex]9-x^{2}-y^{2}\\,\\geq\\,0.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>This inequality can be written in the form<\/p>\n<div style=\"text-align: center;\">[latex]x^{2}+y^{2}\\,\\leq\\,9.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Therefore, the domain of [latex]g\\,(x,\\ y)[\/latex] is [latex]{\\{(x,\\ y)\\in\\mathbb{R}^{2}\\,|\\,x^{2}+y^{2}\\,\\leq\\,9\\}}[\/latex]. The graph of this set of points can be described as a disk of radius [latex]3[\/latex] centered at the origin. The domain includes the boundary circle as shown in the following graph.<\/li>\n<\/ol>\n<div>\n<div id=\"attachment_946\" style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-946\" class=\"size-full wp-image-946\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/27164019\/4-1-2.jpeg\" alt=\"A circle of radius three with center at the origin. The equation x2 + y2 = 9 is given.\" width=\"417\" height=\"422\" \/><\/p>\n<p id=\"caption-attachment-946\" class=\"wp-caption-text\">Figure 2.\u00a0The domain of the function [latex]\\small{g(x,y)=\\sqrt{9-x^{2}-y^{2}}}[\/latex] is a closed disk of radius [latex]\\small{3}[\/latex].<\/p>\n<\/div>\n<\/div>\n<p>To determine the range of [latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}[\/latex] we start with a point [latex](x_0,\\ y_0)[\/latex] on the boundary of the domain, which is defined by the relation [latex]x^{2}+y^{2}=9[\/latex]. It follows that [latex]x^2_0+y^2_0=9[\/latex] and<\/p>\n<div style=\"text-align: center;\">[latex]g\\,(x_0,\\ y_0)=\\sqrt{9-x^2_0-y^2_0}=\\sqrt{9-(x^2_0+y^2_0)}=\\sqrt{9-9}=0[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p>If [latex]x^2_0+y^2_0=0[\/latex] (in other words, [latex]x_0=y_0=0[\/latex]), then<\/p>\n<div style=\"text-align: center;\">[latex]g\\,(x_0,\\ y_0)=\\sqrt{9-x^2_0-y^2_0}=\\sqrt{9-(x^2_0+y^2_0)}=\\sqrt{9-0}=3[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>This is the maximum value of the function. Given any value [latex]c[\/latex] between 0 and 3, we can find an entire set of points inside the domain of [latex]g[\/latex] such that [latex]g\\,(x,\\ y)=c[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc} \\hfill {\\sqrt{9-x^{2}-y^{2}}} & =\\hfill & {c} \\hfill \\\\ \\hfill {9-x^{2}-y^{2}} & =\\hfill & {c^{2}}\\hfill \\\\ \\hfill {x^{2}+y^{2}} & =\\hfill & {9-c^{2}.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Since [latex]9-c^{2}\\,>\\,0[\/latex], this describes a circle of radius [latex]\\sqrt{9-c^{2}}[\/latex] centered at the origin. Any point on this circle satisfies the equation [latex]g\\,(x,\\ y)=c[\/latex]. Therefore, the range of this function can be written in interval notation as [latex][0,\\ 3][\/latex].<\/p>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the domain and range of the function [latex]f\\,(x,\\ y)=\\sqrt{36-9x^{2}-9y^{2}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793933114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793933114\" class=\"hidden-answer\" style=\"display: none\">\nThe domain is the shaded circle defined by the inequality [latex]9x^{2}+9y^{2}\\,\\leq\\,36[\/latex], which\u00a0has a circle of radius [latex]2[\/latex] as its boundary. The range is [latex][0,\\ 6][\/latex].<\/p>\n<div id=\"attachment_991\" style=\"width: 352px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-991\" class=\"size-full wp-image-991\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28231310\/4-1-tryitans1.jpeg\" alt=\"A circle of radius two with center at the origin. The equation x2 + y2 \u2264 4 is given.\" width=\"342\" height=\"347\" \/><\/p>\n<p id=\"caption-attachment-991\" class=\"wp-caption-text\">Figure 3.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186147&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=UaWAuFLTy98&amp;video_target=tpm-plugin-8c5jetnr-UaWAuFLTy98\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.1_transcript.html\">transcript for \u201cCP 4.1\u201d here (opens in new window).<\/a><\/div>\n<h2>Graphing Functions of Two Variables<\/h2>\n<p>Suppose we wish to graph the function [latex]z=(x,\\ y)[\/latex]. This function has two independent variables ([latex]x[\/latex] and [latex]y[\/latex]) and one dependent variable ([latex]z[\/latex]). When graphing a function [latex]y=f\\,(x)[\/latex] of one variable, we use the Cartesian plane. We are able to graph any ordered pair [latex](x,\\ y)[\/latex] in the plane, and every point in the plane has an ordered pair [latex](x,\\ y)[\/latex] associated with it. With a function of two variables, each ordered pair [latex](x,\\ y)[\/latex] in the domain of the function is mapped to a real number [latex]z[\/latex]. Therefore the graph of the function [latex]f[\/latex] consists of ordered pairs [latex](x,\\ y,\\ z)[\/latex]. The graph of a function [latex]z=f\\,(x,\\ y)[\/latex] of two variables is called a\u00a0<strong>surface<\/strong>.<\/p>\n<p>To understand more completely the concept of plotting a set of ordered triples to obtain a surface in three-dimensional space, imagine the [latex]x,\\ y)[\/latex] coordinate system laying flat. Then, every point in the domain of the function [latex]f[\/latex] has a unique [latex]z[\/latex]-value associated with it. If [latex]z[\/latex] is positive, then the graphed point is located above the [latex]xy[\/latex]-plane. The set of all the graphed points becomes the two-dimensional surface that is the graph of the function [latex]f[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing Functions of Two Variables<\/h3>\n<p>Create a graph of each of the following functions:<\/p>\n<ol>\n<li>[latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}[\/latex]<\/li>\n<li>[latex]f\\,(x,\\ y)=x^{2}+y^{2}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1267793933114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1267793933114\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>In the previous example, we determined that the domain of [latex]g\\,(x,\\ y)=\\sqrt{9-x^{2}-y^{2}}[\/latex] is [latex]\\{(x,\\ y)\\in\\mathbb{R}^{2}\\,|\\,x^{2}+y^{2}\\,\\leq\\,9\\}[\/latex] and the range is [latex]{z\\in\\mathbb{R}^{2}\\,|\\,0\\,\\leq\\,z\\,\\leq\\,3}[\/latex]. When [latex]x^{2}+y^{2}=9[\/latex] we have [latex]g\\,(x,\\ y)=0[\/latex]. Therefore, any point on the circle of radius [latex]3[\/latex] centered at the origin in the [latex]xy[\/latex]-plane maps to [latex]z=0[\/latex] in [latex]\\mathbb{R}^3[\/latex]. If [latex]x^{2}+y^{2}=8[\/latex], then [latex]g\\,(x,\\ y)=1[\/latex], so any point on the circle of radius [latex]2\\sqrt{2}[\/latex] centered at the origin in the [latex]xy[\/latex]-plane maps to [latex]z=1[\/latex] in [latex]\\mathbb{R}^{3}[\/latex]. As [latex]x^{2}+y^{2}[\/latex] gets closer to zero, the value of [latex]z[\/latex] approaches [latex]3[\/latex]. When [latex]x^{2}+y^{2}=0[\/latex], then [latex]g\\,(x,\\ y)=3[\/latex]. This is the origin in the [latex]xy[\/latex]-plane. If [latex]x^{2}+y^{2}[\/latex] is equal to any other value between [latex]0[\/latex] and [latex]9[\/latex], then [latex]g\\,(x,\\ y)[\/latex] equals some other constant between [latex]0[\/latex] and [latex]3[\/latex]. The surface described by this function is a hemisphere centered at the origin with radius\u00a0[latex]3[\/latex] as shown in the following graph.\n<div>\n<div id=\"attachment_949\" style=\"width: 607px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-949\" class=\"size-full wp-image-949\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/27164347\/4-1-3.jpeg\" alt=\"A hemisphere with center at the origin. The equation z = g(x, y) = the square root of the quantity (9 \u2013 x2 \u2013 y2) is given.\" width=\"597\" height=\"612\" \/><\/p>\n<p id=\"caption-attachment-949\" class=\"wp-caption-text\">Figure 4. <span class=\"os-caption\">Graph of the hemisphere represented by the given function of two variables.<\/span><\/p>\n<\/div>\n<\/div>\n<\/li>\n<li>This function also contains the expression [latex]x^{2}+y^{2}[\/latex]. Setting this expression equal to various values starting at zero, we obtain circles of increasing radius. The minimum value of [latex]f\\,(x,\\ y)=x^{2}+y^{2}[\/latex] is zero (attained when [latex]x=y=0[\/latex]). When [latex]x=0[\/latex], the function becomes [latex]z=y^{2}[\/latex], and when [latex]y=0[\/latex], then the function becomes [latex]z=x^{2}[\/latex]. These are cross-sections of the graph, and are parabolas. Recall from Introduction to Vectors in Space that the name of the graph of [latex]f\\,(x,\\ y)=x^{2}+y^{2}[\/latex] is a\u00a0<em>paraboloid<\/em>. The graph of [latex]f[\/latex] appears in the following graph.\n<div>\n<div id=\"attachment_951\" style=\"width: 595px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-951\" class=\"size-full wp-image-951\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/27164441\/4-1-4.jpeg\" alt=\"A paraboloid with vertex at the origin. The equation z = f(x, y) = x2 + y2 is given.\" width=\"585\" height=\"574\" \/><\/p>\n<p id=\"caption-attachment-951\" class=\"wp-caption-text\">Figure 5.\u00a0A paraboloid is the graph of the given function of two variables.<\/p>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Nuts and Bolts<\/h3>\n<p>A profit function for a hardware manufacturer is given by<\/p>\n<p style=\"text-align: center;\">[latex]f\\,(x,\\ y)=16-(x-3)^{2}-(y-2)^{2}[\/latex],<\/p>\n<p>where [latex]x[\/latex] is the number of nuts sold per month (measured in thousands) and [latex]y[\/latex] represents the number of bolts sold per month (measured in thousands). Profit is measured in thousands of dollars. Sketch a graph of this function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794933114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794933114\" class=\"hidden-answer\" style=\"display: none\">\nThis function is a polynomial function in two variables. The domain of [latex]f[\/latex] consists of [latex](x,\\ y)[\/latex] coordinate pairs that yield a nonnegative profit:<\/p>\n<div style=\"text-align: center;\">[latex]16-(x-3)^{2}-(y-2)^{2}\\,\\geq\\,0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex](x-3)^{2}-(y-2)^{2}\\,\\leq\\,16.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p style=\"text-align: left;\">This is a disk of radius\u00a0[latex]4[\/latex] centered at [latex](3,\\ 2)[\/latex]. A further restriction is that both [latex]x[\/latex] and [latex]y[\/latex] must be nonnegative. When [latex]x=3[\/latex] and [latex]y=2[\/latex], [latex]f\\,(x,\\ y)=16[\/latex]. Note that it is possible for either value to be a noninteger; for example, it is possible to sell\u00a0[latex]2.5[\/latex] thousand nut in a month. The domain, therefore, contains thousands of points, so we can consider all points within the disk. For any [latex]z<16[\/latex], we can solve the equation [latex]f\\,(x,\\ y)=z[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]16-(x-3)^{2}-(y-2)^{2}=z[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex](x-3)^{2}-(y-2)^{2}=16-z[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p style=\"text-align: left;\">Since [latex]z<16[\/latex], we know that [latex]16-z>0[\/latex], so the previous equation describes a circle with radius [latex]\\sqrt{16-z}[\/latex] centered at the point [latex](3,\\ 2)[\/latex]. Therefore, the range of [latex]f\\,(x,\\ y)[\/latex] is [latex]\\{z\\in\\mathbb{R}\\,|z\\,\\leq\\,16\\}[\/latex]. The graph of [latex]f\\,(x,\\ y)[\/latex] is also a paraboloid, and this paraboloid points downward as shown.<\/p>\n<div id=\"attachment_953\" style=\"width: 506px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-953\" class=\"size-full wp-image-953\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/27164529\/4-1-5.jpeg\" alt=\"A paraboloid center seemingly on the positive z axis. The equation z = f(x, y) = 16 \u2013 (x \u2013 3)2 \u2013 (y \u2013 2)2 is given.\" width=\"496\" height=\"517\" \/><\/p>\n<p id=\"caption-attachment-953\" class=\"wp-caption-text\">Figure 6.\u00a0The graph of the given function of two variables is also a paraboloid.<\/p>\n<\/div>\n<p style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3907\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 4.1. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 4.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3907","chapter","type-chapter","status-publish","hentry"],"part":22,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3907","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":7,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3907\/revisions"}],"predecessor-version":[{"id":5828,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3907\/revisions\/5828"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/22"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3907\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=3907"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=3907"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=3907"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=3907"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}