(*^ ::[ Information = "This is a Mathematica Notebook file. It contains ASCII text, and can be transferred by email, ftp, or other text-file transfer utility. It should be read or edited using a copy of Mathematica or MathReader. If you received this as email, use your mail application or copy/paste to save everything from the line containing (*^ down to the line containing ^*) into a plain text file. On some systems you may have to give the file a name ending with ".ma" to allow Mathematica to recognize it as a Notebook. The line below identifies what version of Mathematica created this file, but it can be opened using any other version as well."; FrontEndVersion = "NeXT Mathematica Notebook Front End Version 2.2"; NeXTStandardFontEncoding; fontset = title, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, bold, L1, e8, 24, "Times"; ; fontset = subtitle, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, bold, L1, e6, 18, "Times"; ; fontset = subsubtitle, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, italic, L1, e6, 14, "Times"; ; fontset = section, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, grayBox, M22, bold, L1, a20, 18, "Times"; ; fontset = subsection, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, blackBox, M19, bold, L1, a15, 14, "Times"; ; fontset = subsubsection, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, whiteBox, M18, bold, L1, a12, 12, "Times"; ; fontset = text, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = smalltext, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 10, "Times"; ; fontset = input, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeInput, M42, N23, bold, L1, 12, "Courier"; ; fontset = output, output, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L-5, 12, "Courier"; ; fontset = message, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = print, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = info, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = postscript, PostScript, formatAsPostScript, output, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeGraphics, M7, l34, w282, h287, L1, 12, "Courier"; ; fontset = name, inactive, noPageBreakInGroup, nohscroll, preserveAspect, M7, italic, B65535, L1, 10, "Times"; ; fontset = header, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, italic, L1, 12, "Times"; ; fontset = leftheader, L0, 12; fontset = footer, inactive, nohscroll, noKeepOnOnePage, preserveAspect, center, M7, italic, L1, 12, "Times"; ; fontset = leftfooter, L0, 12; fontset = help, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = clipboard, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = completions, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12, "Courier"; ; fontset = special1, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special2, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special3, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special4, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special5, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; automaticGrouping; currentKernel; ] :[font = title; inactive; preserveAspect; startGroup] BIGCIRC :[font = section; inactive; preserveAspect; startGroup] BRIEF ABSTRACT :[font = subsection; inactive; preserveAspect; endGroup] We seek to determine the maximum area circle which can be inscribed several selected regions using geometric optimization strategies. :[font = section; inactive; Cclosed; preserveAspect; startGroup] GENERAL INFORMATION :[font = subsection; inactive; preserveAspect] FileName: BIGCIRC :[font = subsection; inactive; preserveAspect] Full title: Inscribe a circle of maximum area in a region bounded above by a function and below by the x-axis. :[font = subsection; inactive; preserveAspect] Last Revision Date: 18 November 1996. :[font = subsection; inactive; preserveAspect] Developer: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. :[font = subsection; inactive; preserveAspect] Contacts: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Aaron D. Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Aaron.Klebanoff@Rose-Hulman.Edu. FAX: 812-877-3198. :[font = subsection; inactive; preserveAspect; endGroup] Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. :[font = section; inactive; Cclosed; preserveAspect; startGroup] STATEMENT OF PROBLEM :[font = subsection; inactive; preserveAspect] (a) Consider the region bounded above by the function f(x) = 9 + 8x - x^2 and below by the x-axis. Determine the largest circle bounded by the region. :[font = subsection; inactive; preserveAspect] (b) Consider the region bounded above by the function f(x) = - x^4/20 + 11/15 x^3 - 17/5 x^2 + 24/5 x + 6 and below by the x-axis. Determine the largest circle bounded by the region. :[font = subsection; inactive; preserveAspect] (c) Consider the region bounded above by the function f(x) = 8 + 6.2 x - 1.7 x^2 + .1 x^3 and below by the x-axis between x = -1 and x = 8. Determine the largest circle bounded by the region. :[font = subsection; inactive; preserveAspect; startGroup] While this would appear to be an abstract only problem consider these two situations. :[font = subsubsection; inactive; preserveAspect] (1) We have purchased a great number of identical scrap plates of sheet metal and we wish to cut out the largest circle bounded by the boundaries of this scrap sheet for making another product. :[font = subsubsection; inactive; preserveAspect; endGroup; endGroup] (2) We have an irregular shaped field and we wish to set up a circular irrigating sprinkler system. How much area can we irrigate, i.e. what is the largest circle we can inscribe in this field, its radius, and its center? :[font = section; inactive; Cclosed; preserveAspect; startGroup] KEYWORDS :[font = subsection; inactive; preserveAspect; endGroup] Geometric optimization, maximum area, inscribed circle, tangent line, system of nonlinear equations. :[font = section; inactive; Cclosed; preserveAspect; startGroup] TEACHER NOTES :[font = subsection; inactive; preserveAspect] ISSUES RELATED TO THE PROBLEM :[font = subsection; inactive; preserveAspect; startGroup] Prerequisites :[font = subsubsection; inactive; preserveAspect; endGroup] Optimization strategies may be useful, point of tangency to a curve, and system of nonlinear equations. :[font = subsection; inactive; preserveAspect; startGroup] Time allotment - time management :[font = subsubsection; inactive; preserveAspect] This could be a big problem and may need some guidance to see how to determine the necessary conditions for a maximum area for the inscribed circle, i.e. the geometric constraints. We suggest permitting 15 minutes for brainstroming, picture drawing, and beginning the initial strategy. :[font = subsubsection; inactive; preserveAspect] If the teacher wishes to suggest the approach, we offer in the Possible Solutions section hints that can be given to move the class along. Setting up of the constraints together with the complexities of solving (if possible) need time. :[font = subsubsection; inactive; preserveAspect; endGroup] It may just be that articulating strategies, going after a few plausible candidates, and offering up some plots will be sufficient for the purposes of understanding what the problem is asking for. These plots may motivate the multiple tangency approach we use in our Possible Solutions. :[font = subsection; inactive; preserveAspect; startGroup] Expectations :[font = subsubsection; inactive; preserveAspect; endGroup] We would expect students to sketch quite a bit before making necessary mathematical conjectures. :[font = subsection; inactive; preserveAspect; startGroup] Future payoffs :[font = subsubsection; inactive; preserveAspect; endGroup] Students will have utilized a number of different calculus skills in doing this problem and reviewed slopes of lines and orthogonal lines. :[font = subsection; inactive; preserveAspect; startGroup] Extensions :[font = subsubsection; inactive; preserveAspect] One can consider other functions which give rise to different geometric configurations. Some quite regular and others quite pathological, e.g., consider transcendental functions. :[font = subsubsection; inactive; preserveAspect; endGroup] One context may be that we have purchased a good bit of scrap metal sheets (dimension in inches say) in the shape of the region described and we wish to salvage what we can. So we seek to cut out the largest circular disk from the material to make platters. :[font = subsection; inactive; preserveAspect; endGroup] References and Sources :[font = section; inactive; Cclosed; preserveAspect; startGroup] POSSIBLE SOLUTIONS :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] (a) Analysis for f(x) = 9 + 8x - x^2. (Using actual distance function.) :[font = subsubsection; inactive; preserveAspect; startGroup] We enter the function and determine where it crosses the axes in order to get a reasonable plot. :[font = input; preserveAspect] f[x_] = 9 - 8x - x^2; :[font = input; preserveAspect; startGroup] solRoots = Solve[f[x]==0,x] //N :[font = output; output; inactive; preserveAspect; endGroup] {{x -> -9.}, {x -> 1.}} ;[o] {{x -> -9.}, {x -> 1.}} :[font = input; preserveAspect] xl = x/.solRoots[[1]]; xr = x/.solRoots[[2]]; :[font = input; preserveAspect; startGroup] fPlot = Plot[f[x],{x,xl,xr},AspectRatio->Automatic] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; preserveAspect] We believe that the circle will be tangent to the curve at two points, say (x0, f(x0)) and (x1, f(x1)) and tangent to the x-axis at say (a, 0). We need to determine these three values: x0, x1, and a. :[font = subsubsection; inactive; preserveAspect; startGroup] We find the lines perpendicular to the curve at the points of tangency and determine where they intersect. Then the distances from each of the tangency points to the center of the inscribed circle should be the same and should be equal to the height from the x-axis. :[font = input; preserveAspect] eq0 = y - f[x0] == -1/f'[x0] (x - x0); :[font = input; preserveAspect] eq1 = y - f[x1] == -1/f'[x1] (x - x1); :[font = input; preserveAspect; endGroup] sol = Expand[Solve[{eq0,eq1},{x,y}]]; :[font = subsubsection; inactive; preserveAspect; startGroup] pt is the point of intersection of the lines perpendicular to the curve at the points of tangency. :[font = input; preserveAspect; endGroup] pt = {x,y}/.sol[[1]]; :[font = subsubsection; inactive; preserveAspect] So we see we have the point of intersection pt = (x, y) in terms of x0 and x1. We need to determine x0 and x1 and hence need two equations in these unknowns. Thus we determine the x0 and x1 values so that (1) the distance from pt to (x1, f(x1)) is the same as the distance from pt to (x0, f(x0)) and (2) the distance from the x axis to pt is the same as the distance from pt to, say (x0, f(x0)). :[font = subsubsection; inactive; preserveAspect; startGroup] (1) leads to equation d1 and (2) leads to equation d2. :[font = input; preserveAspect] d1 = Sqrt[(pt[[1]] - x0)^2 + (pt[[2]] - f[x0])^2] == Sqrt[(pt[[1]] - x1)^2 + (pt[[2]] - f[x1])^2]; :[font = input; preserveAspect; endGroup] d2 = Sqrt[(pt[[1]] - x0)^2 + (pt[[2]] - f[x0])^2] == pt[[2]]; :[font = subsubsection; inactive; preserveAspect; startGroup] We solve for x0 and x1. Be sure to play with a number of starting values in order to get sensible answers which appear reasonable in the diagram. :[font = input; preserveAspect; startGroup] solp = FindRoot[{d1,d2},{x0,-8},{x1,0}, MaxIterations->50] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] {x0 -> -8.47213595499958, x1 -> 0.4721359549995795} ;[o] {x0 -> -8.47214, x1 -> 0.472136} :[font = subsubsection; inactive; preserveAspect; startGroup] So now we have the x coordinates of our two points of tangency (x0, f(x0)) and (x1, f(x1)) on the curve. We call them z0 and z1 respectively. :[font = input; preserveAspect; endGroup] z0 = x0/.solp; z1 = x1/.solp; :[font = subsubsection; inactive; preserveAspect; startGroup] And we determine the point, newpt, which will serve as the center of the circle. :[font = input; preserveAspect; startGroup] newpt = pt/.solp :[font = output; output; inactive; preserveAspect; endGroup; endGroup] {-3.999999999999864, 4.500000000000015} ;[o] {-4., 4.5} :[font = subsubsection; inactive; preserveAspect; startGroup] This means our contact with the x-axis occurs at point (a, 0) where a is given below: :[font = input; preserveAspect; startGroup] a = pt[[1]]/.solp :[font = output; output; inactive; preserveAspect; endGroup; endGroup] -3.999999999999864 ;[o] -4. :[font = subsubsection; inactive; preserveAspect; startGroup] And the radius of the maximum inscribed circle is given by rad below: :[font = input; preserveAspect; startGroup] rad = pt[[2]]/.solp :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 4.500000000000015 ;[o] 4.5 :[font = subsubsection; inactive; preserveAspect; startGroup] We plot the two points of tangency (x0, f(x0)) and (x1, f(x1)) on the curve. :[font = input; preserveAspect; startGroup] pts = Show[Graphics[{PointSize[.1],Point[{z0,f[z0]}], Point[{z1,f[z1]}],Point[{a,0}]}], AspectRatio->Automatic, PlotRange->{{-10,4},{-4,10}}] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; preserveAspect; startGroup] We plot the circle which is tangent to the two points (x0, f(x0)) and (x1, f(x1)) on the curve and the x-axis. :[font = input; preserveAspect; startGroup] circle = ParametricPlot[{rad Cos[t] + a, rad Sin[t] + rad}, {t,0, 2 Pi}, AspectRatio->Automatic, PlotRange->{{-10,4},{-4,10}}] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; preserveAspect; startGroup] And we superimpose all three plots to see the situation. :[font = input; preserveAspect; startGroup] Show[fPlot,pts,circle] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; preserveAspect; startGroup] Finally we determine the area of this maximum circle inscribed in our region. :[font = input; preserveAspect; startGroup] MaxArea = Pi newpt[[2]]^2//N :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup] 63.61725123519375 ;[o] 63.6173 :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] (b) Analysis for f(x) = - x^4/20 + 11/15 x^3 - 17/5 x^2 + 24/5 x + 6. We use a sum squared function (which we explain below) to minimize for solving two distance equations will not work in this case. :[font = subsubsection; inactive; preserveAspect; startGroup] We enter the function and determine where it crosses the axes in order to get a reasonable plot. :[font = input; preserveAspect] f[x_] = - x^4/20 + 11/15 x^3 - 17/5 x^2 + 24/5 x + 6; :[font = input; preserveAspect; startGroup] solRoots = Solve[f[x]==0,x]//N :[font = output; output; inactive; preserveAspect; endGroup] {{x -> 3.842353646794983 - 2.345856074462544*I}, {x -> 3.842353646794983 + 2.345856074462544*I}, {x -> -0.7643670159979146}, {x -> 7.746326389074616}} ;[o] {{x -> 3.84235 - 2.34586 I}, {x -> 3.84235 + 2.34586 I}, {x -> -0.764367}, {x -> 7.74633}} :[font = input; preserveAspect] xl = x/.solRoots[[3]]; xr = x/.solRoots[[4]]; :[font = input; preserveAspect; startGroup] fPlot = Plot[f[x],{x,xl,xr},AspectRatio->Automatic] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; preserveAspect] We believe that the circle will be tangent to the curve at two points, say (x0, f(x0)) and (x1, f(x1)) and tangent to the x-axis at say (a, 0). We need to determine these three values: x0, x1, and a. :[font = subsubsection; inactive; preserveAspect; startGroup] We find the lines perpendicular to the curve at the points of tangency and determine where they intersect. Then the distances from each of the tangency points to the center of the inscribed circle should be the same and should be equal to the height from the x-axis. :[font = input; preserveAspect] eq0 = y - f[x0] == -1/f'[x0] (x - x0); :[font = input; preserveAspect] eq1 = y - f[x1] == -1/f'[x1] (x - x1); :[font = input; preserveAspect; endGroup] sol = Expand[Solve[{eq0,eq1},{x,y}]]; :[font = subsubsection; inactive; preserveAspect; startGroup] pt is the point of intersection of the lines perpendicular to the curve at the points of tangency. :[font = input; preserveAspect; endGroup] pt = {x,y}/.sol[[1]]; :[font = subsubsection; inactive; preserveAspect] So we see we have the point of intersection pt = (x, y) in terms of x0 and x1. We need to determine x0 and x1 and hence need two equations in these unknowns. Thus we determine the x0 and x1 values so that (1) the distance from pt to (x1, f(x1)) is the same as the distance from pt to (x0, f(x0)) and (2) the distance from the x axis to pt is the same as the distance from pt to, say (x0, f(x0)) . :[font = subsubsection; inactive; preserveAspect] Here we square both sides of the above distance equation and substract to obtain expressions in x0 and x1 which must equal zero. :[font = subsubsection; inactive; preserveAspect] Then we square each expression and add them to get a function r(x0,x1) which is equal to 0 if and only if each of the distance equations is satisfied. :[font = subsubsection; inactive; preserveAspect] In order to determine when the non-negative function r(x0,x1), the sum of two squares, is zero we determine where this function has a minimum. :[font = subsubsection; inactive; preserveAspect; startGroup] We do this because Mathematica's FindRoot command for this system of equations resulting from the distance equations did not converge. :[font = input; preserveAspect] d1 = (pt[[1]] - x0)^2 + (pt[[2]] - f[x0])^2 - ((pt[[1]] - x1)^2 + (pt[[2]] - f[x1])^2) == 0; :[font = input; preserveAspect] d2 = (pt[[1]] - x0)^2 + (pt[[2]] - f[x0])^2 - pt[[2]]^2 == 0; :[font = input; preserveAspect] r[x0_,x1_] = d1[[1]]^2 + d2[[1]]^2; :[font = input; preserveAspect; startGroup] solp = FindMinimum[r[x0,x1],{x0,0},{x1,4}, MaxIterations->50] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] {1.987838164590714*10^-19, {x0 -> -0.457193071533683, x1 -> 3.550294661651487}} ;[o] -19 {1.98784 10 , {x0 -> -0.457193, x1 -> 3.55029}} :[font = subsubsection; inactive; preserveAspect; startGroup] So now we have the x coordinates of our two points of tangency (x0, f(x0)) and (x1, f(x1)) on the curve. We call them z0 and z1 respectively. :[font = input; preserveAspect; endGroup] z0 = x0/.solp[[2]]; z1 = x1/.solp[[2]]; :[font = subsubsection; inactive; preserveAspect; startGroup] And we determine the point, newpt, which will serve as the center of the circle. :[font = input; preserveAspect; startGroup] newpt = pt/.solp[[2]] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] {2.226388348670793, 2.70258542484294} ;[o] {2.22639, 2.70259} :[font = subsubsection; inactive; preserveAspect; startGroup] This means our contact with the x-axis occurs at point (a, 0) where a is given below: :[font = input; preserveAspect; startGroup] a = pt[[1]]/.solp[[2]] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 2.226388348670793 ;[o] 2.22639 :[font = subsubsection; inactive; preserveAspect; startGroup] And the radius of the maximum inscribed circle is given by rad below: :[font = input; preserveAspect; startGroup] rad = pt[[2]]/.solp[[2]] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 2.70258542484294 ;[o] 2.70259 :[font = subsubsection; inactive; preserveAspect; startGroup] We plot the two points of tangency (x0, f(x0)) and (x1, f(x1)) on the curve. :[font = input; preserveAspect; startGroup] pts = Show[Graphics[{PointSize[.10],Point[{z0,f[z0]}], Point[{z1,f[z1]}]}], AspectRatio->Automatic, PlotRange->{{-4,10},{-4,10}}] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; preserveAspect; startGroup] We plot the circle which is tangent to the two points (x0, f(x0)) and (x1, f(x1)) on the curve and the x-axis. :[font = input; preserveAspect; startGroup] circle = ParametricPlot[{rad Cos[t] + a, rad Sin[t] + rad}, {t,0, 2 Pi},AspectRatio->Automatic, PlotRange->{{-4,10},{-4,10}}] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; preserveAspect; startGroup] And we superimpose all three plots to see the situation. :[font = input; preserveAspect; startGroup] Show[fPlot,pts,circle] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; preserveAspect; startGroup] Finally we determine the area of this maximum circle inscribed in our region. :[font = input; preserveAspect; startGroup] MaxArea = Pi newpt[[2]]^2//N :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup] 22.94609214354158 ;[o] 22.9461 :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] (c) Analysis for f(x) = 8 + 6.2 x - 1.7 x^2 + .1 x^3. Using actual distance function. :[font = subsubsection; inactive; preserveAspect; startGroup] We enter the function and determine where it crosses the axes in order to get a reasonable plot. :[font = input; preserveAspect] f[x_] = 8. + 6.2 x - 1.7 x^2 + 0.1 x^3; :[font = input; preserveAspect; startGroup] solRoots = Solve[f[x]==0,x]//N :[font = output; output; inactive; preserveAspect; endGroup] {{x -> -1.}, {x -> 7.999999999999997}, {x -> 10.}} ;[o] {{x -> -1.}, {x -> 8.}, {x -> 10.}} :[font = input; preserveAspect] xl = x/.solRoots[[1]]; xr = x/.solRoots[[2]]; :[font = input; preserveAspect; startGroup] fPlot = Plot[f[x],{x,xl,xr},AspectRatio->Automatic] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; preserveAspect] We believe that the circle will be tangent to the curve at two points, say (x0, f(x0)) and (x1, f(x1)) and tangent to the x-axis at say (a, 0). We need to determine these three values: x0, x1, and a. :[font = subsubsection; inactive; preserveAspect; startGroup] We find the lines perpendicular to the curve at the points of tangency and determine where they intersect. Then the distances from each of the tangency points to the center of the inscribed circle should be the same and should be equal to the height from the x-axis. :[font = input; preserveAspect] eq0 = y - f[x0] == -1/f'[x0] (x - x0); :[font = input; preserveAspect] eq1 = y - f[x1] == -1/f'[x1] (x - x1); :[font = input; preserveAspect; endGroup] sol = Expand[Solve[{eq0,eq1},{x,y}]]; :[font = subsubsection; inactive; preserveAspect; startGroup] pt is the point of intersection of the lines perpendicular to the curve at the points of tangency. :[font = input; preserveAspect; endGroup] pt = {x,y}/.sol[[1]]; :[font = subsubsection; inactive; preserveAspect] So we see we have the point of intersection pt = (x, y) in terms of x0 and x1. We need to determine x0 and x1 and hence need two equations in these unknowns. Thus we determine the x0 and x1 values so that (1) the distance from pt to (x1, f(x1)) is the same as the distance from pt to (x0, f(x0)) and (2) the distance from the x axis to pt is the same as the distance from pt to, say (x0, f(x0)). :[font = subsubsection; inactive; preserveAspect; startGroup] (1) leads to equation d1 and (2) leads to equation d2. :[font = input; preserveAspect] d1 = Sqrt[(pt[[1]] - x0)^2 + (pt[[2]] - f[x0])^2] == Sqrt[(pt[[1]] - x1)^2 + (pt[[2]] - f[x1])^2]; :[font = input; preserveAspect; endGroup] d2 = Sqrt[(pt[[1]] - x0)^2 + (pt[[2]] - f[x0])^2] == pt[[2]]; :[font = subsubsection; inactive; preserveAspect; startGroup] We solve for x0 and x1. :[font = input; preserveAspect; startGroup] solp = FindRoot[{d1,d2},{x0,1},{x1,7}, MaxIterations->50] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] {x0 -> -0.562959815105306, x1 -> 6.313467398697935} ;[o] {x0 -> -0.56296, x1 -> 6.31347} :[font = subsubsection; inactive; preserveAspect; startGroup] So now we have the x coordinates of our two points of tangency (x0, f(x0)) and (x1, f(x1)) on the curve. We call them z0 and z1 respectively. :[font = input; preserveAspect; endGroup] z0 = x0/.solp; z1 = x1/.solp; :[font = subsubsection; inactive; preserveAspect; startGroup] And we determine the point, newpt, which will serve as the center of the circle. :[font = input; preserveAspect; startGroup] newpt = pt/.solp :[font = output; output; inactive; preserveAspect; endGroup; endGroup] {2.937758030127267, 3.526595773447527} ;[o] {2.93776, 3.5266} :[font = subsubsection; inactive; preserveAspect; startGroup] This means our contact with the x-axis occurs at point (a, 0) where a is given below: :[font = input; preserveAspect; startGroup] a = pt[[1]]/.solp :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 2.937758030127267 ;[o] 2.93776 :[font = subsubsection; inactive; preserveAspect; startGroup] And the radius of the maximum inscribed circle is given by rad below: :[font = input; preserveAspect; startGroup] rad = pt[[2]]/.solp :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 3.526595773447527 ;[o] 3.5266 :[font = subsubsection; inactive; preserveAspect; startGroup] We plot the two points of tangency (x0, f(x0)) and (x1, f(x1)) on the curve. :[font = input; preserveAspect; startGroup] pts = Show[Graphics[{PointSize[.10],Point[{z0,f[z0]}], Point[{z1,f[z1]}]}], AspectRatio->Automatic, PlotRange->{{-4,10},{-4,10}}] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; preserveAspect; startGroup] We plot the circle which is tangent to the two points (x0, f(x0)) and (x1, f(x1)) on the curve and the x-axis. :[font = input; preserveAspect; startGroup] circle = ParametricPlot[{rad Cos[t] + a, rad Sin[t] + rad}, {t,0, 2 Pi},AspectRatio->Automatic, PlotRange->{{-4,10},{-4,10}}] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; preserveAspect; startGroup] And we superimpose all three plots to see the situation. :[font = input; preserveAspect; startGroup] Show[fPlot,pts,circle] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; preserveAspect; startGroup] Finally we determine the area of this maximum circle inscribed in our region. :[font = input; preserveAspect; startGroup] MaxArea = Pi rad^2//N :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup; endGroup] 39.07160377078884 ;[o] 39.0716 :[font = section; inactive; Cclosed; preserveAspect; startGroup] ISSUES IN SOLUTION :[font = subsection; inactive; preserveAspect; endGroup; endGroup] The first problem and the last problem are similar in nature and the second problem needed another solution strategy for numerically obtaining the points. ^*)