{
  "index": "1938-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "2. A can buoy is to be made of three pieces, namely, a cylinder and two equal cones, the altitude of each cone being equal to the altitude of the cylinder. For a given area of surface, what shape will have the greatest volume?",
  "solution": "Solution. Let \\( r \\) be the radius of the cylinder, and \\( h \\) its altitude. The given condition is\n\\[\nS=2 \\pi r h+2\\left(\\pi r \\sqrt{h^{2}+r^{2}}\\right)=\\text { constant }\n\\]\nand the volume of the buoy is\n\\[\nV=\\pi r^{2} h+\\frac{2 \\pi r^{2} h}{3}=\\frac{5 \\pi r^{2} h}{3}\n\\]\n\nThe required problem is to find the maximum value of \\( V \\) subject to condition (1). This can be done by the method of Lagrange multipliers, but in this particular problem it is easier to solve (1) for \\( h \\) and express \\( V \\) as a function of \\( r \\). We have\n\\[\n(S-2 \\pi r h)^{2}=4 \\pi^{2} r^{2}\\left(h^{2}+r^{2}\\right)\n\\]\nwhence\n\\[\nh=\\frac{S^{2}-4 \\pi^{2} r^{4}}{4 \\pi r S}\n\\]\nand the expression for \\( V \\) becomes\n\\[\nV=\\frac{5 r}{12 S}\\left(S^{2}-4 \\pi^{2} r^{4}\\right)\n\\]\n\nSince \\( r \\) and \\( V \\) must be positive, the domain of interest is given by\n\\[\n0<r<\\sqrt[4]{S^{2} / 4 \\pi^{2}}\n\\]\n\nWe compute the derivative and equate it to zero to get\n\\[\n\\frac{d V}{d r}=\\frac{5 S}{12}-\\frac{100 \\pi^{2} r^{4}}{12 S}=0 .\n\\]\n\nThe only critical value is\n\\[\nr_{0}=\\sqrt[4]{\\frac{S^{2}}{20 \\pi^{2}}}\n\\]\n\nSince \\( V \\rightarrow 0 \\) as \\( r \\rightarrow 0 \\) or as \\( r \\rightarrow \\sqrt[4]{S^{2} / 4 \\pi^{2}} \\), and is positive in between, the critical value \\( r_{0} \\) yields a maximum for \\( V \\).\n\nThe corresponding value of \\( h \\) is found from (3) to be \\( h_{0}=\\frac{2}{5} \\sqrt{5} r_{0} \\). The shape of the buoy is completely determined by the ratio\n\\[\n\\frac{h_{0}}{r_{0}}=\\frac{2}{5} \\sqrt{5}\n\\]",
  "vars": [
    "r",
    "h",
    "V",
    "r_0",
    "h_0"
  ],
  "params": [
    "S"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "r": "radius",
        "h": "altitude",
        "V": "volume",
        "r_0": "criticalradius",
        "h_0": "criticalaltitude",
        "S": "surfacearea"
      },
      "question": "2. A can buoy is to be made of three pieces, namely, a cylinder and two equal cones, the altitude of each cone being equal to the altitude of the cylinder. For a given area of surface, what shape will have the greatest volume?",
      "solution": "Solution. Let \\( radius \\) be the radius of the cylinder, and \\( altitude \\) its altitude. The given condition is\n\\[\nsurfacearea = 2 \\pi radius altitude + 2\\left(\\pi radius \\sqrt{altitude^{2}+radius^{2}}\\right)=\\text { constant }\n\\]\nand the volume of the buoy is\n\\[\nvolume = \\pi radius^{2} altitude + \\frac{2 \\pi radius^{2} altitude}{3} = \\frac{5 \\pi radius^{2} altitude}{3}\n\\]\n\nThe required problem is to find the maximum value of \\( volume \\) subject to condition (1). This can be done by the method of Lagrange multipliers, but in this particular problem it is easier to solve (1) for \\( altitude \\) and express \\( volume \\) as a function of \\( radius \\). We have\n\\[\n(surfacearea - 2 \\pi radius altitude)^{2}=4 \\pi^{2} radius^{2}\\left(altitude^{2}+radius^{2}\\right)\n\\]\nwhence\n\\[\naltitude = \\frac{surfacearea^{2}-4 \\pi^{2} radius^{4}}{4 \\pi radius surfacearea}\n\\]\nand the expression for \\( volume \\) becomes\n\\[\nvolume = \\frac{5 radius}{12 surfacearea}\\left(surfacearea^{2}-4 \\pi^{2} radius^{4}\\right)\n\\]\n\nSince \\( radius \\) and \\( volume \\) must be positive, the domain of interest is given by\n\\[\n0< radius < \\sqrt[4]{surfacearea^{2} / 4 \\pi^{2}}\n\\]\n\nWe compute the derivative and equate it to zero to get\n\\[\n\\frac{d volume}{d radius} = \\frac{5 surfacearea}{12} - \\frac{100 \\pi^{2} radius^{4}}{12 surfacearea} = 0 .\n\\]\n\nThe only critical value is\n\\[\ncriticalradius = \\sqrt[4]{\\frac{surfacearea^{2}}{20 \\pi^{2}}}\n\\]\n\nSince \\( volume \\rightarrow 0 \\) as \\( radius \\rightarrow 0 \\) or as \\( radius \\rightarrow \\sqrt[4]{surfacearea^{2} / 4 \\pi^{2}} \\), and is positive in between, the critical value \\( criticalradius \\) yields a maximum for \\( volume \\).\n\nThe corresponding value of \\( altitude \\) is found from (3) to be \\( criticalaltitude = \\frac{2}{5} \\sqrt{5} criticalradius \\). The shape of the buoy is completely determined by the ratio\n\\[\n\\frac{criticalaltitude}{criticalradius} = \\frac{2}{5} \\sqrt{5}\n\\]\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "r": "monument",
        "h": "daybreak",
        "V": "calendar",
        "r_0": "monumental",
        "h_0": "daybreaker",
        "S": "landscape"
      },
      "question": "2. A can buoy is to be made of three pieces, namely, a cylinder and two equal cones, the altitude of each cone being equal to the altitude of the cylinder. For a given area of surface, what shape will have the greatest volume?",
      "solution": "Solution. Let \\( monument \\) be the radius of the cylinder, and \\( daybreak \\) its altitude. The given condition is\n\\[\nlandscape=2 \\pi monument daybreak+2\\left(\\pi monument \\sqrt{daybreak^{2}+monument^{2}}\\right)=\\text { constant }\n\\]\nand the volume of the buoy is\n\\[\ncalendar=\\pi monument^{2} daybreak+\\frac{2 \\pi monument^{2} daybreak}{3}=\\frac{5 \\pi monument^{2} daybreak}{3}\n\\]\n\nThe required problem is to find the maximum value of \\( calendar \\) subject to condition (1). This can be done by the method of Lagrange multipliers, but in this particular problem it is easier to solve (1) for \\( daybreak \\) and express \\( calendar \\) as a function of \\( monument \\). We have\n\\[\n(landscape-2 \\pi monument daybreak)^{2}=4 \\pi^{2} monument^{2}\\left(daybreak^{2}+monument^{2}\\right)\n\\]\nwhence\n\\[\ndaybreak=\\frac{landscape^{2}-4 \\pi^{2} monument^{4}}{4 \\pi monument landscape}\n\\]\nand the expression for \\( calendar \\) becomes\n\\[\ncalendar=\\frac{5 monument}{12 landscape}\\left(landscape^{2}-4 \\pi^{2} monument^{4}\\right)\n\\]\n\nSince \\( monument \\) and \\( calendar \\) must be positive, the domain of interest is given by\n\\[\n0<monument<\\sqrt[4]{landscape^{2} / 4 \\pi^{2}}\n\\]\n\nWe compute the derivative and equate it to zero to get\n\\[\n\\frac{d calendar}{d monument}=\\frac{5 landscape}{12}-\\frac{100 \\pi^{2} monument^{4}}{12 landscape}=0 .\n\\]\n\nThe only critical value is\n\\[\nmonumental=\\sqrt[4]{\\frac{landscape^{2}}{20 \\pi^{2}}}\n\\]\n\nSince \\( calendar \\rightarrow 0 \\) as \\( monument \\rightarrow 0 \\) or as \\( monument \\rightarrow \\sqrt[4]{landscape^{2} / 4 \\pi^{2}} \\), and is positive in between, the critical value \\( monumental \\) yields a maximum for \\( calendar \\).\n\nThe corresponding value of \\( daybreak \\) is found from (3) to be \\( daybreaker=\\frac{2}{5} \\sqrt{5} monumental \\). The shape of the buoy is completely determined by the ratio\n\\[\n\\frac{daybreaker}{monumental}=\\frac{2}{5} \\sqrt{5}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "r": "perimeterlength",
        "h": "depthvalue",
        "V": "surfacearea",
        "r_0": "minimumdepth",
        "h_0": "maximumperimeter",
        "S": "corevolume"
      },
      "question": "2. A can buoy is to be made of three pieces, namely, a cylinder and two equal cones, the altitude of each cone being equal to the altitude of the cylinder. For a given area of surface, what shape will have the greatest volume?",
      "solution": "Solution. Let \\( perimeterlength \\) be the radius of the cylinder, and \\( depthvalue \\) its altitude. The given condition is\n\\[\ncorevolume = 2 \\pi perimeterlength depthvalue + 2\\left(\\pi perimeterlength \\sqrt{depthvalue^{2}+perimeterlength^{2}}\\right)=\\text { constant }\n\\]\nand the volume of the buoy is\n\\[\nsurfacearea = \\pi perimeterlength^{2} depthvalue + \\frac{2 \\pi perimeterlength^{2} depthvalue}{3}=\\frac{5 \\pi perimeterlength^{2} depthvalue}{3}\n\\]\n\nThe required problem is to find the maximum value of \\( surfacearea \\) subject to condition (1). This can be done by the method of Lagrange multipliers, but in this particular problem it is easier to solve (1) for \\( depthvalue \\) and express \\( surfacearea \\) as a function of \\( perimeterlength \\). We have\n\\[\n(corevolume-2 \\pi perimeterlength depthvalue)^{2}=4 \\pi^{2} perimeterlength^{2}\\left(depthvalue^{2}+perimeterlength^{2}\\right)\n\\]\nwhence\n\\[\ndepthvalue = \\frac{corevolume^{2}-4 \\pi^{2} perimeterlength^{4}}{4 \\pi perimeterlength corevolume}\n\\]\nand the expression for \\( surfacearea \\) becomes\n\\[\nsurfacearea = \\frac{5 perimeterlength}{12 corevolume}\\left(corevolume^{2}-4 \\pi^{2} perimeterlength^{4}\\right)\n\\]\n\nSince \\( perimeterlength \\) and \\( surfacearea \\) must be positive, the domain of interest is given by\n\\[\n0<perimeterlength<\\sqrt[4]{corevolume^{2} / 4 \\pi^{2}}\n\\]\n\nWe compute the derivative and equate it to zero to get\n\\[\n\\frac{d surfacearea}{d perimeterlength}=\\frac{5 corevolume}{12}-\\frac{100 \\pi^{2} perimeterlength^{4}}{12 corevolume}=0 .\n\\]\n\nThe only critical value is\n\\[\nminimumdepth=\\sqrt[4]{\\frac{corevolume^{2}}{20 \\pi^{2}}}\n\\]\n\nSince \\( surfacearea \\rightarrow 0 \\) as \\( perimeterlength \\rightarrow 0 \\) or as \\( perimeterlength \\rightarrow \\sqrt[4]{corevolume^{2} / 4 \\pi^{2}} \\), and is positive in between, the critical value \\( minimumdepth \\) yields a maximum for \\( surfacearea \\).\n\nThe corresponding value of \\( depthvalue \\) is found from (3) to be \\( maximumperimeter = \\frac{2}{5} \\sqrt{5} minimumdepth \\). The shape of the buoy is completely determined by the ratio\n\\[\n\\frac{maximumperimeter}{minimumdepth}=\\frac{2}{5} \\sqrt{5}\n\\]"
    },
    "garbled_string": {
      "map": {
        "r": "qzxwvtnp",
        "h": "yrklsfhd",
        "V": "mnbvcxza",
        "r_0": "ploikmnj",
        "h_0": "ujhytgrf",
        "S": "asdfghjk"
      },
      "question": "2. A can buoy is to be made of three pieces, namely, a cylinder and two equal cones, the altitude of each cone being equal to the altitude of the cylinder. For a given area of surface, what shape will have the greatest volume?",
      "solution": "Solution. Let \\( qzxwvtnp \\) be the radius of the cylinder, and \\( yrklsfhd \\) its altitude. The given condition is\n\\[\nasdfghjk=2 \\pi qzxwvtnp yrklsfhd+2\\left(\\pi qzxwvtnp \\sqrt{yrklsfhd^{2}+qzxwvtnp^{2}}\\right)=\\text { constant }\n\\]\nand the volume of the buoy is\n\\[\nmnbvcxza=\\pi qzxwvtnp^{2} yrklsfhd+\\frac{2 \\pi qzxwvtnp^{2} yrklsfhd}{3}=\\frac{5 \\pi qzxwvtnp^{2} yrklsfhd}{3}\n\\]\n\nThe required problem is to find the maximum value of \\( mnbvcxza \\) subject to condition (1). This can be done by the method of Lagrange multipliers, but in this particular problem it is easier to solve (1) for \\( yrklsfhd \\) and express \\( mnbvcxza \\) as a function of \\( qzxwvtnp \\). We have\n\\[\n(asdfghjk-2 \\pi qzxwvtnp yrklsfhd)^{2}=4 \\pi^{2} qzxwvtnp^{2}\\left(yrklsfhd^{2}+qzxwvtnp^{2}\\right)\n\\]\nwhence\n\\[\nyrklsfhd=\\frac{asdfghjk^{2}-4 \\pi^{2} qzxwvtnp^{4}}{4 \\pi qzxwvtnp asdfghjk}\n\\]\nand the expression for \\( mnbvcxza \\) becomes\n\\[\nmnbvcxza=\\frac{5 qzxwvtnp}{12 asdfghjk}\\left(asdfghjk^{2}-4 \\pi^{2} qzxwvtnp^{4}\\right)\n\\]\n\nSince \\( qzxwvtnp \\) and \\( mnbvcxza \\) must be positive, the domain of interest is given by\n\\[\n0<qzxwvtnp<\\sqrt[4]{asdfghjk^{2} / 4 \\pi^{2}}\n\\]\n\nWe compute the derivative and equate it to zero to get\n\\[\n\\frac{d mnbvcxza}{d qzxwvtnp}=\\frac{5 asdfghjk}{12}-\\frac{100 \\pi^{2} qzxwvtnp^{4}}{12 asdfghjk}=0 .\n\\]\n\nThe only critical value is\n\\[\nploikmnj=\\sqrt[4]{\\frac{asdfghjk^{2}}{20 \\pi^{2}}}\n\\]\n\nSince \\( mnbvcxza \\rightarrow 0 \\) as \\( qzxwvtnp \\rightarrow 0 \\) or as \\( qzxwvtnp \\rightarrow \\sqrt[4]{asdfghjk^{2} / 4 \\pi^{2}} \\), and is positive in between, the critical value \\( ploikmnj \\) yields a maximum for \\( mnbvcxza \\).\n\nThe corresponding value of \\( yrklsfhd \\) is found from (3) to be \\( ujhytgrf=\\frac{2}{5} \\sqrt{5} ploikmnj \\). The shape of the buoy is completely determined by the ratio\n\\[\n\\frac{ujhytgrf}{ploikmnj}=\\frac{2}{5} \\sqrt{5}\n\\]"
    },
    "kernel_variant": {
      "question": "A float is to be built from two solid pieces: a right circular cylinder of radius \\(r\\) and altitude \\(h\\), and (attached to its top) a right circular cone that has the same base radius \\(r\\) and altitude \\(\\tfrac{h}{2}\\).  \n\nAll of the outside surface – that is, the lateral area of the cylinder, the lateral area of the cone, and the exposed circular bottom of the cylinder – is to be coated with exactly a fixed amount \\(S\\) of waterproof paint. (The circular interface where the cone meets the cylinder is internal and receives no paint.)  \n\nFor this fixed paint supply, determine the value of the ratio \\(h/r\\) that maximises the volume enclosed by the float.  (Give the exact algebraic condition that this ratio satisfies and a numerical approximation.)",
      "solution": "Let k = h/r (>0) be the desired ratio.  Express every quantity in terms of r and k.\n\n1.  Painted area\n   S = πr² + 2πr(k r) + πr√[r² + (k r/2)²]\n     = πr² + 2πk r² + πr²√(1 + k²/4)\n     = πr² F(k),            where F(k) := 1 + 2k + √(1 + k²/4).\n\n2.  From this, with S fixed,\n   r = √[S /(πF(k))].\n\n3.  Volume\n   V = πr²(k r) + (1/3)πr²(k r/2)\n     = πk r³ + (1/6)πk r³\n     = (7/6)πk r³\n     = (7/6)π [S /(πF(k))]^{3/2} k\n     = constant · G(k)\n   with G(k) := k / F(k)^{3/2}.\n\n   Maximising V is therefore equivalent to maximising G(k).\n\n4.  Set g(k) = ln G(k) = ln k – (3/2) ln F(k).  Then\n      g'(k) = 1/k – (3/2)·F'(k)/F(k) = 0.\n\n   Compute  F'(k) = 2 + k/[4√(1 + k²/4)].  Setting g'(k)=0 gives\n      2/k = 3F'(k)/F(k).\n   Substituting F and F' and clearing the square root yields\n      15k³ – 32k² + 96k – 128 = 0.        (∗)\n\n5.  Polynomial (∗) has exactly one positive root.  Numerically one finds\n      k_max = h/r ≈ 1.55198  (to five significant figures).\n\n6.  End‐point check: as k→0⁺ or k→∞, G(k)→0, so the critical point furnished by (∗) indeed gives the absolute maximum of the volume for the prescribed paint area.\n\nThus the cylinder should be about 1.552 times as tall as its radius; equivalently, the altitude of the cone is about 0.776 r.\n\nExact condition:  15(h/r)³ − 32(h/r)² + 96(h/r) − 128 = 0.",
      "_meta": {
        "core_steps": [
          "Express surface-area constraint S(r,h) and volume V(r,h) from geometry",
          "Solve the constraint for h (or use a Lagrange multiplier) to get V=V(r) alone",
          "Differentiate V(r), set dV/dr = 0, locate admissible critical r",
          "Check endpoints to confirm the critical point yields the maximum",
          "Translate that r into the optimal h/r shape ratio"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "How many identical cones are attached to the cylinder",
            "original": 2
          },
          "slot2": {
            "description": "Altitude of each cone as a multiple of the cylinder's altitude",
            "original": 1
          },
          "slot3": {
            "description": "Whether the flat circular bases are counted in the fixed surface area",
            "original": "not counted (only lateral areas used)"
          },
          "slot4": {
            "description": "Which quantity is held fixed vs. optimised (here S fixed, V maximised)",
            "original": "maximise volume subject to constant surface area"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}