{"id":192,"date":"2021-07-30T17:29:28","date_gmt":"2021-07-30T17:29:28","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=192"},"modified":"2022-10-29T02:19:00","modified_gmt":"2022-10-29T02:19:00","slug":"putting-it-together-differentiation-of-functions-of-several-variables","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/putting-it-together-differentiation-of-functions-of-several-variables\/","title":{"raw":"Putting It Together: Differentiation of Functions of Several Variables","rendered":"Putting It Together: Differentiation of Functions of Several Variables"},"content":{"raw":"

Profitable Golf Balls<\/h2>\r\n[caption id=\"attachment_229\" align=\"aligncenter\" width=\"450\"]\"A Figure 1.[\/caption]\r\n\r\nPro-[latex]T[\/latex] company has developed a profit model that depends on the number [latex]x[\/latex] of golf balls sold per month (measured in thousands), and the number of hours per month of advertising [latex]y[\/latex], according to the function\r\n

[latex]z=f(x,y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex]<\/p>\r\nwhere [latex]z[\/latex]\u00a0is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold is\u00a0[latex]50,000[\/latex],\u00a0and the maximum number of hours of advertising that can be purchased is\u00a0[latex]25[\/latex]. Find the values of\u00a0[latex]x[\/latex]\u00a0and\u00a0[latex]y[\/latex]\u00a0that maximize profit, and find the maximum profit.\r\n

Solution<\/span><\/h4>\r\nUsing the problem-solving strategy, step 1\u00a0involves finding the critical points of [latex]f[\/latex]\u00a0on its domain. Therefore, we first calculate [latex]f_{x}(x,y)[\/latex]\u00a0and\u00a0[latex]f_{y}(x,y)[\/latex],\u00a0then set them each equal to zero:\r\n

[latex]\\begin{array}{c}f_{x}(x,y) &=& 48-2x-2y \\hfill \\\\f_{y}(x,y) &=& 96-2x-18y \\hfill \\end{array}[\/latex]<\/p>\r\nSetting them equal to zero yields the system of equations\r\n

[latex]\\begin{array}{c}48-2x-2y &=& 0 \\hfill \\\\96-2x-18y &=& 0 \\hfill \\end{array}[\/latex]<\/p>\r\nThe solution to this system is [latex]x=21[\/latex] and [latex]y=3[\/latex]. Therefore [latex](21,3)[\/latex]\u00a0is a critical point of [latex]f[\/latex]. Calculating [latex]f(21,3)[\/latex] gives\u00a0[latex]f(21,3)=48(21)+96(3)-21^{2}-2(21)(3)-9(3)^{2}=648[\/latex].\r\n\r\nThe domain of this function is [latex]0\\le{x}\\le{50}[\/latex] and [latex]0\\le{y}\\le{25}[\/latex]\u00a0as shown in the following graph.\r\n\r\n[caption id=\"attachment_324\" align=\"aligncenter\" width=\"307\"]\"A Figure 2. Graph of the domain of the function[latex]z=f(x,y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex].[\/caption][latex]L_{1}[\/latex]\u00a0is the line segment connecting [latex](0,0)[\/latex] and\u00a0[latex](50,0)[\/latex],\u00a0and it can be parameterized by the equations [latex]x(t)=t[\/latex], [latex]y(t)=0[\/latex] for [latex]0\\le{t}\\le{50}[\/latex].\u00a0We then define [latex]g(t)=f(x(t),y(t))[\/latex]:\r\n

[latex]\\begin{array}{c}g(t) &=&f(x(t),y(t)) \\hfill \\\\ &=& f(t,0) \\hfill \\\\ &=& 48t+96(0)-y^{2}-2(t)(0)-9(0)^{2} \\hfill \\\\ &=& 48t-t^{2} \\hfill \\end{array}[\/latex]<\/p>\r\nSetting [latex]g^{\\prime}(t)=0[\/latex] yields the critical point [latex]t=24[\/latex],\u00a0which corresponds to the point [latex](24,0)[\/latex]\u00a0in the domain of [latex]f[\/latex]. Calculating [latex]f(24,0)[\/latex] gives [latex]576[\/latex].\r\n\r\n[latex]L_{2}[\/latex]\u00a0is the line segment connecting [latex](50,0)[\/latex] and\u00a0[latex](50,25)[\/latex],\u00a0and it can be parameterized by the equations [latex]x(t)=50[\/latex], [latex]y(t)=t[\/latex] for [latex]0\\le{t}\\le{25}[\/latex].\u00a0Once again we define [latex]g(t)=f(x(t),y(t))[\/latex]:\r\n

[latex]\\begin{array}{c}g(t) &=&f(x(t),y(t)) \\hfill \\\\ &=& f(50,t) \\hfill \\\\ &=& 48(50)+96t-50^{2}-2(50)t-9t^{2} \\hfill \\\\ &=& -9t^{2}-4t-100 \\hfill \\end{array}[\/latex]<\/p>\r\nThis function has a critical point [latex]t=-\\frac{2}{9}[\/latex],\u00a0which corresponds to the point [latex](50,-\\frac{2}{9})[\/latex].\u00a0This point is not in the domain of\u00a0[latex]f[\/latex].\r\n\r\n[latex]L_{3}[\/latex]\u00a0is the line segment connecting [latex](0,25)[\/latex] and\u00a0[latex](50,25)[\/latex],\u00a0and it can be parameterized by the equations\u00a0[latex]x(t)=t[\/latex], [latex]y(t)=25[\/latex] for [latex]0\\le{t}\\le{50}[\/latex]. We define [latex]g(t)=f(x(t),y(t))[\/latex]:\r\n

[latex]\\begin{array}{c}g(t) &=&f(x(t),y(t)) \\hfill \\\\ &=& f(t,25) \\hfill \\\\ &=& 48(t)+96(25)-t^{2}-2t(25)-9(25)^{2} \\hfill \\\\ &=& -t^{2}-2t-3225 \\hfill \\end{array}[\/latex]<\/p>\r\nThis function has a critical point [latex]t=-1[\/latex],\u00a0which corresponds to the point [latex](-1,25)[\/latex], which is not in the domain.\r\n\r\n[latex]L_{4}[\/latex]\u00a0is the line segment connecting [latex](0)[\/latex] and\u00a0[latex](0,25)[\/latex],\u00a0and it can be parameterized by the equations\u00a0[latex]x(t)=0[\/latex], [latex]y(t)=t[\/latex] for [latex]0\\le{t}\\le{25}[\/latex]. We define [latex]g(t)=f(x(t),y(t))[\/latex]:\r\n

[latex]\\begin{array}{c}g(t) &=&f(x(t),y(t)) \\hfill \\\\ &=& f(0,t) \\hfill \\\\ &=& 48(0)+96t-(0)^{2}-2(0)t-9t^{2} \\hfill \\\\ &=& 96t-t^{2} \\hfill \\end{array}[\/latex]<\/p>\r\nThis function has a critical point [latex]t=\\frac{16}{3}[\/latex],\u00a0which corresponds to the point [latex](0,\\frac{16}{3})[\/latex], which is on the boundary of the domain. Calculating [latex]f(0,\\frac{16}{3})[\/latex] gives [latex]256[\/latex].\r\n\r\nWe also need to find the values of\u00a0<\/span>[latex]f(x,y)[\/latex]\u00a0at the corners of its domain. These corners are located at [latex](0,0),(50,0),(50,25),\\text{ and }(0,25)[\/latex]:\r\n

[latex]\\begin{array}{c}f(0,0) &=& 48(0)+96(0)-(0)^{2}-2(0)(0)-9(0)^{2}=0 \\hfill \\\\f(50,0) &=& 48(50)+96(0)-(50)^{2}-2(50)(0)-9(0)^{2}=-100 \\hfill \\\\f(50,25) &=& 48(50)+96(25)-(500)^{2}-2(50)(25)-9(25)^{2}=-5825 \\hfill \\\\f(0,25) &=& 48(0)+96(25)-(0)^{2}-2(0)(25)-9(25)^{2}=-3225 \\hfill \\end{array}[\/latex]<\/p>\r\nThe maximum critical value is [latex]648[\/latex]\u00a0which occurs at [latex](21,3)[\/latex].\u00a0Therefore, a maximum profit of [latex]\\$648,000[\/latex]\u00a0is realized when [latex]21,000[\/latex]\u00a0golf balls are sold and [latex]33[\/latex] hours of advertising are purchased per month as shown in the following figure.\r\n\r\n[caption id=\"attachment_326\" align=\"aligncenter\" width=\"682\"]\"The Figure 3. The profit function [latex]f(x,y)[\/latex] has a maximum at [latex](21,3,648)[\/latex].[\/caption]","rendered":"

Profitable Golf Balls<\/h2>\n
\"A<\/p>\n

Figure 1.<\/p>\n<\/div>\n

Pro-[latex]T[\/latex] company has developed a profit model that depends on the number [latex]x[\/latex] of golf balls sold per month (measured in thousands), and the number of hours per month of advertising [latex]y[\/latex], according to the function<\/p>\n

[latex]z=f(x,y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex]<\/p>\n

where [latex]z[\/latex]\u00a0is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold is\u00a0[latex]50,000[\/latex],\u00a0and the maximum number of hours of advertising that can be purchased is\u00a0[latex]25[\/latex]. Find the values of\u00a0[latex]x[\/latex]\u00a0and\u00a0[latex]y[\/latex]\u00a0that maximize profit, and find the maximum profit.<\/p>\n

Solution<\/span><\/h4>\n

Using the problem-solving strategy, step 1\u00a0involves finding the critical points of [latex]f[\/latex]\u00a0on its domain. Therefore, we first calculate [latex]f_{x}(x,y)[\/latex]\u00a0and\u00a0[latex]f_{y}(x,y)[\/latex],\u00a0then set them each equal to zero:<\/p>\n

[latex]\\begin{array}{c}f_{x}(x,y) &=& 48-2x-2y \\hfill \\\\f_{y}(x,y) &=& 96-2x-18y \\hfill \\end{array}[\/latex]<\/p>\n

Setting them equal to zero yields the system of equations<\/p>\n

[latex]\\begin{array}{c}48-2x-2y &=& 0 \\hfill \\\\96-2x-18y &=& 0 \\hfill \\end{array}[\/latex]<\/p>\n

The solution to this system is [latex]x=21[\/latex] and [latex]y=3[\/latex]. Therefore [latex](21,3)[\/latex]\u00a0is a critical point of [latex]f[\/latex]. Calculating [latex]f(21,3)[\/latex] gives\u00a0[latex]f(21,3)=48(21)+96(3)-21^{2}-2(21)(3)-9(3)^{2}=648[\/latex].<\/p>\n

The domain of this function is [latex]0\\le{x}\\le{50}[\/latex] and [latex]0\\le{y}\\le{25}[\/latex]\u00a0as shown in the following graph.<\/p>\n

\"A<\/p>\n

Figure 2. Graph of the domain of the function[latex]z=f(x,y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex].<\/p>\n<\/div>\n

[latex]L_{1}[\/latex]\u00a0is the line segment connecting [latex](0,0)[\/latex] and\u00a0[latex](50,0)[\/latex],\u00a0and it can be parameterized by the equations [latex]x(t)=t[\/latex], [latex]y(t)=0[\/latex] for [latex]0\\le{t}\\le{50}[\/latex].\u00a0We then define [latex]g(t)=f(x(t),y(t))[\/latex]:<\/p>\n

[latex]\\begin{array}{c}g(t) &=&f(x(t),y(t)) \\hfill \\\\ &=& f(t,0) \\hfill \\\\ &=& 48t+96(0)-y^{2}-2(t)(0)-9(0)^{2} \\hfill \\\\ &=& 48t-t^{2} \\hfill \\end{array}[\/latex]<\/p>\n

Setting [latex]g^{\\prime}(t)=0[\/latex] yields the critical point [latex]t=24[\/latex],\u00a0which corresponds to the point [latex](24,0)[\/latex]\u00a0in the domain of [latex]f[\/latex]. Calculating [latex]f(24,0)[\/latex] gives [latex]576[\/latex].<\/p>\n

[latex]L_{2}[\/latex]\u00a0is the line segment connecting [latex](50,0)[\/latex] and\u00a0[latex](50,25)[\/latex],\u00a0and it can be parameterized by the equations [latex]x(t)=50[\/latex], [latex]y(t)=t[\/latex] for [latex]0\\le{t}\\le{25}[\/latex].\u00a0Once again we define [latex]g(t)=f(x(t),y(t))[\/latex]:<\/p>\n

[latex]\\begin{array}{c}g(t) &=&f(x(t),y(t)) \\hfill \\\\ &=& f(50,t) \\hfill \\\\ &=& 48(50)+96t-50^{2}-2(50)t-9t^{2} \\hfill \\\\ &=& -9t^{2}-4t-100 \\hfill \\end{array}[\/latex]<\/p>\n

This function has a critical point [latex]t=-\\frac{2}{9}[\/latex],\u00a0which corresponds to the point [latex](50,-\\frac{2}{9})[\/latex].\u00a0This point is not in the domain of\u00a0[latex]f[\/latex].<\/p>\n

[latex]L_{3}[\/latex]\u00a0is the line segment connecting [latex](0,25)[\/latex] and\u00a0[latex](50,25)[\/latex],\u00a0and it can be parameterized by the equations\u00a0[latex]x(t)=t[\/latex], [latex]y(t)=25[\/latex] for [latex]0\\le{t}\\le{50}[\/latex]. We define [latex]g(t)=f(x(t),y(t))[\/latex]:<\/p>\n

[latex]\\begin{array}{c}g(t) &=&f(x(t),y(t)) \\hfill \\\\ &=& f(t,25) \\hfill \\\\ &=& 48(t)+96(25)-t^{2}-2t(25)-9(25)^{2} \\hfill \\\\ &=& -t^{2}-2t-3225 \\hfill \\end{array}[\/latex]<\/p>\n

This function has a critical point [latex]t=-1[\/latex],\u00a0which corresponds to the point [latex](-1,25)[\/latex], which is not in the domain.<\/p>\n

[latex]L_{4}[\/latex]\u00a0is the line segment connecting [latex](0)[\/latex] and\u00a0[latex](0,25)[\/latex],\u00a0and it can be parameterized by the equations\u00a0[latex]x(t)=0[\/latex], [latex]y(t)=t[\/latex] for [latex]0\\le{t}\\le{25}[\/latex]. We define [latex]g(t)=f(x(t),y(t))[\/latex]:<\/p>\n

[latex]\\begin{array}{c}g(t) &=&f(x(t),y(t)) \\hfill \\\\ &=& f(0,t) \\hfill \\\\ &=& 48(0)+96t-(0)^{2}-2(0)t-9t^{2} \\hfill \\\\ &=& 96t-t^{2} \\hfill \\end{array}[\/latex]<\/p>\n

This function has a critical point [latex]t=\\frac{16}{3}[\/latex],\u00a0which corresponds to the point [latex](0,\\frac{16}{3})[\/latex], which is on the boundary of the domain. Calculating [latex]f(0,\\frac{16}{3})[\/latex] gives [latex]256[\/latex].<\/p>\n

We also need to find the values of\u00a0<\/span>[latex]f(x,y)[\/latex]\u00a0at the corners of its domain. These corners are located at [latex](0,0),(50,0),(50,25),\\text{ and }(0,25)[\/latex]:<\/p>\n

[latex]\\begin{array}{c}f(0,0) &=& 48(0)+96(0)-(0)^{2}-2(0)(0)-9(0)^{2}=0 \\hfill \\\\f(50,0) &=& 48(50)+96(0)-(50)^{2}-2(50)(0)-9(0)^{2}=-100 \\hfill \\\\f(50,25) &=& 48(50)+96(25)-(500)^{2}-2(50)(25)-9(25)^{2}=-5825 \\hfill \\\\f(0,25) &=& 48(0)+96(25)-(0)^{2}-2(0)(25)-9(25)^{2}=-3225 \\hfill \\end{array}[\/latex]<\/p>\n

The maximum critical value is [latex]648[\/latex]\u00a0which occurs at [latex](21,3)[\/latex].\u00a0Therefore, a maximum profit of [latex]\\$648,000[\/latex]\u00a0is realized when [latex]21,000[\/latex]\u00a0golf balls are sold and [latex]33[\/latex] hours of advertising are purchased per month as shown in the following figure.<\/p>\n

\"The<\/p>\n

Figure 3. The profit function [latex]f(x,y)[\/latex] has a maximum at [latex](21,3,648)[\/latex].<\/p>\n<\/div>\n\n\t\t\t

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