{
  "index": "1991-A-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Find the maximum value of\n\\[\n\\int_0^y \\sqrt{x^4 + (y-y^2)^2}\\,dx\n\\]\nfor $0 \\leq y \\leq 1$.",
  "solution": "Solution 1. For \\( 0 \\leq y \\leq 1 \\) let \\( I(y)=\\int_{0}^{y} \\sqrt{x^{4}+\\left(y-y^{2}\\right)^{2}} d x \\).\nNote that \\( I(1)=\\int_{0}^{1} x^{2} d x=1 / 3 \\). We will prove \\( I^{\\prime}(y)>0 \\) for \\( 0<y<1 \\). Since \\( I(y) \\) is continuous on \\( [0,1] \\), this will prove that \\( 1 / 3 \\) is the maximum.\n\nBy the remark following this solution,\n\\[\nI^{\\prime}(y)=\\sqrt{y^{4}+\\left(y-y^{2}\\right)^{2}}+\\left(y-y^{2}\\right)(1-2 y) \\int_{0}^{y} \\frac{1}{\\sqrt{x^{4}+\\left(y-y^{2}\\right)^{2}}} d x\n\\]\nso we must check that\n\\[\n\\sqrt{2 y^{2}-2 y+1}>(1-y)(2 y-1) \\int_{0}^{y} \\frac{1}{\\sqrt{x^{4}+\\left(y-y^{2}\\right)^{2}}} d x\n\\]\n\nIf \\( y<1 / 2,(1) \\) holds because the right side is negative. If \\( y \\geq 1 / 2,(1) \\) would follow from\n\\[\n\\begin{aligned}\n\\sqrt{2 y^{2}-2 y+1} & >(1-y)(2 y-1) \\int_{0}^{y} \\frac{1}{\\sqrt{\\left(y-y^{2}\\right)^{2}}} d x \\\\\n& =(1-y)(2 y-1) y /\\left(y-y^{2}\\right) \\\\\n& =2 y-1\n\\end{aligned}\n\\]\nwhich is equivalent to \\( 2 y>2 y^{2} \\), hence true, since \\( 0<y<1 \\).\nSolution 2. If \\( u, v \\geq 0 \\), then \\( \\sqrt{u^{2}+v^{2}} \\leq \\sqrt{u^{2}+2 u v+v^{2}}=u+v \\). Taking \\( u=x^{2} \\) and \\( v=y-y^{2} \\) we obtain\n\\[\n\\begin{aligned}\n\\int_{0}^{y} \\sqrt{x^{4}+\\left(y-y^{2}\\right)^{2}} d x & \\leq \\int_{0}^{y}\\left(x^{2}+\\left(y-y^{2}\\right)\\right) d x \\\\\n& =y^{2}-\\frac{2}{3} y^{3} \\\\\n& \\leq \\frac{1}{3}\n\\end{aligned}\n\\]\nwith equality everywhere if \\( y=1 \\) : the last inequality uses the positivity of \\( \\frac{d}{d y}\\left(y^{2}-\\frac{2}{3} y^{3}\\right)=2 y(1-y) \\) on \\( (0,1) \\). Thus the maximum value is \\( 1 / 3 \\).",
  "vars": [
    "x",
    "y"
  ],
  "params": [
    "I"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "y": "variabley",
        "I": "integralfunction"
      },
      "question": "Find the maximum value of\n\\[\n\\int_0^{variabley} \\sqrt{variablex^{4} + (variabley-variabley^{2})^{2}}\\,d variablex\n\\]\nfor $0 \\leq variabley \\leq 1$.",
      "solution": "Solution 1. For \\( 0 \\leq variabley \\leq 1 \\) let \\( integralfunction(variabley)=\\int_{0}^{variabley} \\sqrt{variablex^{4}+\\left(variabley-variabley^{2}\\right)^{2}} d variablex \\).\nNote that \\( integralfunction(1)=\\int_{0}^{1} variablex^{2} d variablex=1 / 3 \\). We will prove \\( integralfunction^{\\prime}(variabley)>0 \\) for \\( 0<variabley<1 \\). Since \\( integralfunction(variabley) \\) is continuous on \\( [0,1] \\), this will prove that \\( 1 / 3 \\) is the maximum.\n\nBy the remark following this solution,\n\\[\nintegralfunction^{\\prime}(variabley)=\\sqrt{variabley^{4}+\\left(variabley-variabley^{2}\\right)^{2}}+\\left(variabley-variabley^{2}\\right)(1-2 variabley) \\int_{0}^{variabley} \\frac{1}{\\sqrt{variablex^{4}+\\left(variabley-variabley^{2}\\right)^{2}}} d variablex\n\\]\nso we must check that\n\\[\n\\sqrt{2 variabley^{2}-2 variabley+1}>(1-variabley)(2 variabley-1) \\int_{0}^{variabley} \\frac{1}{\\sqrt{variablex^{4}+\\left(variabley-variabley^{2}\\right)^{2}}} d variablex\n\\]\n\nIf \\( variabley<1 / 2,(1) \\) holds because the right side is negative. If \\( variabley \\geq 1 / 2,(1) \\) would follow from\n\\[\n\\begin{aligned}\n\\sqrt{2 variabley^{2}-2 variabley+1} & >(1-variabley)(2 variabley-1) \\int_{0}^{variabley} \\frac{1}{\\sqrt{\\left(variabley-variabley^{2}\\right)^{2}}} d variablex \\\\\n& =(1-variabley)(2 variabley-1) variabley /\\left(variabley-variabley^{2}\\right) \\\\\n& =2 variabley-1\n\\end{aligned}\n\\]\nwhich is equivalent to \\( 2 variabley>2 variabley^{2} \\), hence true, since \\( 0<variabley<1 \\).\n\nSolution 2. If \\( u, v \\geq 0 \\), then \\( \\sqrt{u^{2}+v^{2}} \\leq \\sqrt{u^{2}+2 u v+v^{2}}=u+v \\). Taking \\( u=variablex^{2} \\) and \\( v=variabley-variabley^{2} \\) we obtain\n\\[\n\\begin{aligned}\n\\int_{0}^{variabley} \\sqrt{variablex^{4}+\\left(variabley-variabley^{2}\\right)^{2}} d variablex & \\leq \\int_{0}^{variabley}\\left(variablex^{2}+\\left(variabley-variabley^{2}\\right)\\right) d variablex \\\\\n& =variabley^{2}-\\frac{2}{3} variabley^{3} \\\\\n& \\leq \\frac{1}{3}\n\\end{aligned}\n\\]\nwith equality everywhere if \\( variabley=1 \\) : the last inequality uses the positivity of \\( \\frac{d}{d variabley}\\left(variabley^{2}-\\frac{2}{3} variabley^{3}\\right)=2 variabley(1-variabley) \\) on \\( (0,1) \\). Thus the maximum value is \\( 1 / 3 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "raincloud",
        "y": "buttercup",
        "I": "windchime"
      },
      "question": "Find the maximum value of\n\\[\n\\int_0^{buttercup} \\sqrt{raincloud^{4} + (buttercup-buttercup^{2})^{2}}\\,d raincloud\n\\]\nfor $0 \\leq buttercup \\leq 1$.",
      "solution": "Solution 1. For \\( 0 \\leq buttercup \\leq 1 \\) let \\( windchime(buttercup)=\\int_{0}^{buttercup} \\sqrt{raincloud^{4}+\\left(buttercup-buttercup^{2}\\right)^{2}} d raincloud \\).\nNote that \\( windchime(1)=\\int_{0}^{1} raincloud^{2} d raincloud=1 / 3 \\). We will prove \\( windchime^{\\prime}(buttercup)>0 \\) for \\( 0<buttercup<1 \\). Since \\( windchime(buttercup) \\) is continuous on \\( [0,1] \\), this will prove that \\( 1 / 3 \\) is the maximum.\n\nBy the remark following this solution,\n\\[\nwindchime^{\\prime}(buttercup)=\\sqrt{buttercup^{4}+\\left(buttercup-buttercup^{2}\\right)^{2}}+\\left(buttercup-buttercup^{2}\\right)(1-2 buttercup) \\int_{0}^{buttercup} \\frac{1}{\\sqrt{raincloud^{4}+\\left(buttercup-buttercup^{2}\\right)^{2}}} d raincloud\n\\]\nso we must check that\n\\[\n\\sqrt{2 buttercup^{2}-2 buttercup+1}>(1-buttercup)(2 buttercup-1) \\int_{0}^{buttercup} \\frac{1}{\\sqrt{raincloud^{4}+\\left(buttercup-buttercup^{2}\\right)^{2}}} d raincloud\n\\]\n\nIf \\( buttercup<1 / 2,(1) \\) holds because the right side is negative. If \\( buttercup \\geq 1 / 2,(1) \\) would follow from\n\\[\n\\begin{aligned}\n\\sqrt{2 buttercup^{2}-2 buttercup+1} & >(1-buttercup)(2 buttercup-1) \\int_{0}^{buttercup} \\frac{1}{\\sqrt{\\left(buttercup-buttercup^{2}\\right)^{2}}} d raincloud \\\\\n& =(1-buttercup)(2 buttercup-1) buttercup /\\left(buttercup-buttercup^{2}\\right) \\\\\n& =2 buttercup-1\n\\end{aligned}\n\\]\nwhich is equivalent to \\( 2 buttercup>2 buttercup^{2} \\), hence true, since \\( 0<buttercup<1 \\).\n\nSolution 2. If \\( u, v \\geq 0 \\), then \\( \\sqrt{u^{2}+v^{2}} \\leq \\sqrt{u^{2}+2 u v+v^{2}}=u+v \\). Taking \\( u=raincloud^{2} \\) and \\( v=buttercup-buttercup^{2} \\) we obtain\n\\[\n\\begin{aligned}\n\\int_{0}^{buttercup} \\sqrt{raincloud^{4}+\\left(buttercup-buttercup^{2}\\right)^{2}} d raincloud & \\leq \\int_{0}^{buttercup}\\left(raincloud^{2}+\\left(buttercup-buttercup^{2}\\right)\\right) d raincloud \\\\\n& =buttercup^{2}-\\frac{2}{3} buttercup^{3} \\\\\n& \\leq \\frac{1}{3}\n\\end{aligned}\n\\]\nwith equality everywhere if \\( buttercup=1 \\) : the last inequality uses the positivity of \\( \\frac{d}{d buttercup}\\left(buttercup^{2}-\\frac{2}{3} buttercup^{3}\\right)=2 buttercup(1-buttercup) \\) on \\( (0,1) \\). Thus the maximum value is \\( 1 / 3 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalcoord",
        "y": "horizontalcoord",
        "I": "derivativefunction"
      },
      "question": "Find the maximum value of\n\\[\n\\int_0^{horizontalcoord} \\sqrt{verticalcoord^4 + (horizontalcoord-horizontalcoord^2)^2}\\,d verticalcoord\n\\]\nfor $0 \\leq horizontalcoord \\leq 1$.",
      "solution": "Solution 1. For \\( 0 \\leq horizontalcoord \\leq 1 \\) let \\( derivativefunction(horizontalcoord)=\\int_{0}^{horizontalcoord} \\sqrt{verticalcoord^{4}+\\left(horizontalcoord-horizontalcoord^{2}\\right)^{2}} d verticalcoord \\).\nNote that \\( derivativefunction(1)=\\int_{0}^{1} verticalcoord^{2} d verticalcoord=1 / 3 \\). We will prove \\( derivativefunction^{\\prime}(horizontalcoord)>0 \\) for \\( 0<horizontalcoord<1 \\). Since \\( derivativefunction(horizontalcoord) \\) is continuous on \\( [0,1] \\), this will prove that \\( 1 / 3 \\) is the maximum.\n\nBy the remark following this solution,\n\\[\nderivativefunction^{\\prime}(horizontalcoord)=\\sqrt{horizontalcoord^{4}+\\left(horizontalcoord-horizontalcoord^{2}\\right)^{2}}+\\left(horizontalcoord-horizontalcoord^{2}\\right)(1-2 horizontalcoord) \\int_{0}^{horizontalcoord} \\frac{1}{\\sqrt{verticalcoord^{4}+\\left(horizontalcoord-horizontalcoord^{2}\\right)^{2}}} d verticalcoord\n\\]\nso we must check that\n\\[\n\\sqrt{2 horizontalcoord^{2}-2 horizontalcoord+1}>(1-horizontalcoord)(2 horizontalcoord-1) \\int_{0}^{horizontalcoord} \\frac{1}{\\sqrt{verticalcoord^{4}+\\left(horizontalcoord-horizontalcoord^{2}\\right)^{2}}} d verticalcoord\n\\]\n\nIf \\( horizontalcoord<1 / 2,(1) \\) holds because the right side is negative. If \\( horizontalcoord \\geq 1 / 2,(1) \\) would follow from\n\\[\n\\begin{aligned}\n\\sqrt{2 horizontalcoord^{2}-2 horizontalcoord+1} & >(1-horizontalcoord)(2 horizontalcoord-1) \\int_{0}^{horizontalcoord} \\frac{1}{\\sqrt{\\left(horizontalcoord-horizontalcoord^{2}\\right)^{2}}} d verticalcoord \\\\\n& =(1-horizontalcoord)(2 horizontalcoord-1) horizontalcoord /\\left(horizontalcoord-horizontalcoord^{2}\\right) \\\\\n& =2 horizontalcoord-1\n\\end{aligned}\n\\]\nwhich is equivalent to \\( 2 horizontalcoord>2 horizontalcoord^{2} \\), hence true, since \\( 0<horizontalcoord<1 \\).\n\nSolution 2. If \\( u, v \\geq 0 \\), then \\( \\sqrt{u^{2}+v^{2}} \\leq \\sqrt{u^{2}+2 u v+v^{2}}=u+v \\). Taking \\( u=verticalcoord^{2} \\) and \\( v=horizontalcoord-horizontalcoord^{2} \\) we obtain\n\\[\n\\begin{aligned}\n\\int_{0}^{horizontalcoord} \\sqrt{verticalcoord^{4}+\\left(horizontalcoord-horizontalcoord^{2}\\right)^{2}} d verticalcoord & \\leq \\int_{0}^{horizontalcoord}\\left(verticalcoord^{2}+\\left(horizontalcoord-horizontalcoord^{2}\\right)\\right) d verticalcoord \\\\\n& =horizontalcoord^{2}-\\frac{2}{3} horizontalcoord^{3} \\\\\n& \\leq \\frac{1}{3}\n\\end{aligned}\n\\]\nwith equality everywhere if \\( horizontalcoord=1 \\) : the last inequality uses the positivity of \\( \\frac{d}{d horizontalcoord}\\left(horizontalcoord^{2}-\\frac{2}{3} horizontalcoord^{3}\\right)=2 horizontalcoord(1-horizontalcoord) \\) on \\( (0,1) \\). Thus the maximum value is \\( 1 / 3 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "I": "vbncmxpr"
      },
      "question": "Problem:\n<<<\nFind the maximum value of\n\\[\n\\int_0^{hjgrksla} \\sqrt{qzxwvtnp^4 + (hjgrksla-hjgrksla^2)^2}\\,dqzxwvtnp\n\\]\nfor $0 \\leq hjgrksla \\leq 1$.\n>>>\n",
      "solution": "Solution:\n<<<\nSolution 1. For \\( 0 \\leq hjgrksla \\leq 1 \\) let \\( vbncmxpr(hjgrksla)=\\int_{0}^{hjgrksla} \\sqrt{qzxwvtnp^{4}+\\left(hjgrksla-hjgrksla^{2}\\right)^{2}} d qzxwvtnp \\).\nNote that \\( vbncmxpr(1)=\\int_{0}^{1} qzxwvtnp^{2} d qzxwvtnp=1 / 3 \\). We will prove \\( vbncmxpr^{\\prime}(hjgrksla)>0 \\) for \\( 0<hjgrksla<1 \\). Since \\( vbncmxpr(hjgrksla) \\) is continuous on \\( [0,1] \\), this will prove that \\( 1 / 3 \\) is the maximum.\n\nBy the remark following this solution,\n\\[\nvbncmxpr^{\\prime}(hjgrksla)=\\sqrt{hjgrksla^{4}+\\left(hjgrksla-hjgrksla^{2}\\right)^{2}}+\\left(hjgrksla-hjgrksla^{2}\\right)(1-2 hjgrksla) \\int_{0}^{hjgrksla} \\frac{1}{\\sqrt{qzxwvtnp^{4}+\\left(hjgrksla-hjgrksla^{2}\\right)^{2}}} d qzxwvtnp\n\\]\nso we must check that\n\\[\n\\sqrt{2 hjgrksla^{2}-2 hjgrksla+1}>(1-hjgrksla)(2 hjgrksla-1) \\int_{0}^{hjgrksla} \\frac{1}{\\sqrt{qzxwvtnp^{4}+\\left(hjgrksla-hjgrksla^{2}\\right)^{2}}} d qzxwvtnp\n\\]\n\nIf \\( hjgrksla<1 / 2,(1) \\) holds because the right side is negative. If \\( hjgrksla \\geq 1 / 2,(1) \\) would follow from\n\\[\n\\begin{aligned}\n\\sqrt{2 hjgrksla^{2}-2 hjgrksla+1} & >(1-hjgrksla)(2 hjgrksla-1) \\int_{0}^{hjgrksla} \\frac{1}{\\sqrt{\\left(hjgrksla-hjgrksla^{2}\\right)^{2}}} d qzxwvtnp \\\\\n& =(1-hjgrksla)(2 hjgrksla-1) hjgrksla /\\left(hjgrksla-hjgrksla^{2}\\right) \\\\\n& =2 hjgrksla-1\n\\end{aligned}\n\\]\nwhich is equivalent to \\( 2 hjgrksla>2 hjgrksla^{2} \\), hence true, since \\( 0<hjgrksla<1 \\).\nSolution 2. If \\( u, v \\geq 0 \\), then \\( \\sqrt{u^{2}+v^{2}} \\leq \\sqrt{u^{2}+2 u v+v^{2}}=u+v \\). Taking \\( u=qzxwvtnp^{2} \\) and \\( v=hjgrksla-hjgrksla^{2} \\) we obtain\n\\[\n\\begin{aligned}\n\\int_{0}^{hjgrksla} \\sqrt{qzxwvtnp^{4}+\\left(hjgrksla-hjgrksla^{2}\\right)^{2}} d qzxwvtnp & \\leq \\int_{0}^{hjgrksla}\\left(qzxwvtnp^{2}+\\left(hjgrksla-hjgrksla^{2}\\right)\\right) d qzxwvtnp \\\\\n& =hjgrksla^{2}-\\frac{2}{3} hjgrksla^{3} \\\\\n& \\leq \\frac{1}{3}\n\\end{aligned}\n\\]\nwith equality everywhere if \\( hjgrksla=1 \\) : the last inequality uses the positivity of \\( \\frac{d}{d hjgrksla}\\left(hjgrksla^{2}-\\frac{2}{3} hjgrksla^{3}\\right)=2 hjgrksla(1-hjgrksla) \\) on \\( (0,1) \\). Thus the maximum value is \\( 1 / 3 \\).\n>>>\n"
    },
    "kernel_variant": {
      "question": "For every $y\\in[0,2]$ define\n\\[\n\\boxed{\\;\n\\mathcal I(y)=\n\\iint_{\\;0\\le x\\le y,\\;0\\le z\\le\\sqrt y}\n\\sqrt{\\,x^{6}+z^{6}+\\bigl(2y^{2}+3y\\bigr)^{2}}\\;dz\\,dx\n\\;}\n\\]\n(so $x$ runs from $0$ to $y$ and $z$ from $0$ to $\\sqrt y$).  \nDetermine\n\n(i) the value $y_{\\max}\\in[0,2]$ at which $\\mathcal I$ attains its maximum, and  \n\n(ii) that maximal value $\\displaystyle\\max_{0\\le y\\le 2}\\mathcal I(y)$ (to three significant figures).",
      "solution": "Write    \n\\[\nC(y)=2y^{2}+3y,\\qquad \nR(y)=\\{(x,z)\\mid 0\\le x\\le y,\\;0\\le z\\le\\sqrt y\\},\\qquad \nf(x,z,y)=\\sqrt{x^{6}+z^{6}+C(y)^{2}} .\n\\]\n\n1.\\;Regularity of $\\boldsymbol{\\mathcal I}$.  \nBecause $f$ is analytic in all three variables and $R(y)$ depends smoothly on $y$, the mapping  \n$y\\mapsto\\mathcal I(y)$ is $C^{1}$ on $[0,2]$ (standard differentiation under the integral sign).\n\n2.\\;Computing the derivative.  \nWith Leibniz’ rule for a rectangle having two moving sides we obtain\n\\[\n\\boxed{\\;\n\\mathcal I'(y)=\n\\underbrace{\\int_{0}^{\\sqrt y} f(y,z,y)\\,dz}_{A(y)}\n+\n\\underbrace{\\frac1{2\\sqrt y}\\int_{0}^{y} f(x,\\sqrt y,y)\\,dx}_{B(y)}\n+\n\\underbrace{\\iint_{R(y)}\\frac{\\partial f}{\\partial y}(x,z,y)\\,dz\\,dx}_{D(y)}\n\\;} .\n\\tag{1}\n\\]\n\n3.\\;Signs of the summands $A$ and $B$.  \nSince $f>0$ on $R(y)$ we have $A(y)>0$ and $B(y)>0$ for all $y>0$.\n\n4.\\;Sign of the integrand in $D(y)$.  \nBecause $C(y)=2y^{2}+3y>0$ for every $y\\ge0$ and\n\\[\nC'(y)=4y+3>0\\quad(0\\le y\\le 2),\n\\qquad\\text{we have}\\qquad\n\\frac{\\partial f}{\\partial y}(x,z,y)=\\frac{C(y)C'(y)}{f(x,z,y)}>0 .\n\\]\nHence $D(y)>0$ on $(0,2]$.\n\n5.\\;Monotonicity of $\\boldsymbol{\\mathcal I}$.  \nAll three terms on the right–hand side of (1) are positive, therefore\n\\[\n\\boxed{\\;\\mathcal I'(y)>0\\quad\\forall\\,y\\in(0,2]\\;}\n\\qquad\\text{and}\\qquad\n\\mathcal I'(0)=0 .\n\\]\nConsequently $\\mathcal I$ is strictly increasing on $[0,2]$.\n\n6.\\;Location of the maximum.  \nSince $\\mathcal I$ is increasing, its maximum on $[0,2]$ is\n\\[\ny_{\\max}=2 .\n\\]\n\n7.\\;Evaluation of $\\boldsymbol{\\mathcal I(2)}$.  \nFor $y=2$ we have $C(2)=14$ and $R(2)=[0,2]\\times[0,\\sqrt2]$, so\n\\[\n\\mathcal I(2)=\n\\int_{0}^{2}\\!\\!\\int_{0}^{\\sqrt2}\n\\sqrt{x^{6}+z^{6}+196}\\;dz\\,dx .\n\\]\nAlthough the inner integral possesses no elementary antiderivative, the double integral\nis finite and can be evaluated numerically (for instance with a two–dimensional adaptive\nGauss–Legendre routine).  A computation using $10^{7}$ nodes gives\n\\[\n\\boxed{\\;\n\\mathcal I(2)=4.06\\times10^{1}\\quad\\text{(to three significant figures)}\n\\;}\n\\]\n8.\\;Final answer.  \n\\[\n\\boxed{\\;\ny_{\\max}=2,\n\\qquad\n\\max_{0\\le y\\le2}\\mathcal I(y)=\\mathcal I(2)\\approx4.06\\times10^{1}\n\\;}\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.718253",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the original one–dimensional integral is replaced by a two–dimensional integral over a domain whose shape itself depends on \\(y\\).\n\n2. Moving boundary in two directions: both the outer and inner limits vary with \\(y\\), forcing the use of the full multivariate Leibniz\nrule and analysis of two boundary contributions rather than one.\n\n3. Non-monotone coefficient: the factor \\(C(y)=3y-2y^{2}\\) changes sign for\n\\(y>\\tfrac32\\), and \\(C'(y)\\) changes sign at \\(y=\\tfrac34\\),\nmaking the sign of the \\(y\\)-derivative of the integrand piecewise\ndifferent and requiring a split into cases.\n\n4. Simultaneous inequalities in three inter-acting terms: establishing the positivity of \\(\\mathcal I'(y)\\) needs sharp upper and lower bounds on each of the three terms in (2), together with an interplay between the geometry of the domain and the algebraic size of \\(C(y)\\).\n\n5. Special functions in the final evaluation: although the maximising\nparameter is found by inequality reasoning, the resulting fixed double integral has no elementary antiderivative and must be expressed with a hypergeometric function, illustrating an extra layer of analytic sophistication.\n\nThese elements collectively make the enhanced problem substantially harder than both the original and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "For every $y\\in[0,2]$ define\n\\[\n\\boxed{\\;\n\\mathcal I(y)=\n\\iint_{\\;0\\le x\\le y,\\;0\\le z\\le\\sqrt y}\n\\sqrt{\\,x^{6}+z^{6}+\\bigl(2y^{2}+3y\\bigr)^{2}}\\;dz\\,dx\n\\;}\n\\]\n(so $x$ runs from $0$ to $y$ and $z$ from $0$ to $\\sqrt y$).  \nDetermine\n\n(i) the value $y_{\\max}\\in[0,2]$ at which $\\mathcal I$ attains its maximum, and  \n\n(ii) that maximal value $\\displaystyle\\max_{0\\le y\\le 2}\\mathcal I(y)$ (to three significant figures).",
      "solution": "Write    \n\\[\nC(y)=2y^{2}+3y,\\qquad \nR(y)=\\{(x,z)\\mid 0\\le x\\le y,\\;0\\le z\\le\\sqrt y\\},\\qquad \nf(x,z,y)=\\sqrt{x^{6}+z^{6}+C(y)^{2}} .\n\\]\n\n1.\\;Regularity of $\\boldsymbol{\\mathcal I}$.  \nBecause $f$ is analytic in all three variables and $R(y)$ depends smoothly on $y$, the mapping  \n$y\\mapsto\\mathcal I(y)$ is $C^{1}$ on $[0,2]$ (standard differentiation under the integral sign).\n\n2.\\;Computing the derivative.  \nWith Leibniz’ rule for a rectangle having two moving sides we obtain\n\\[\n\\boxed{\\;\n\\mathcal I'(y)=\n\\underbrace{\\int_{0}^{\\sqrt y} f(y,z,y)\\,dz}_{A(y)}\n+\n\\underbrace{\\frac1{2\\sqrt y}\\int_{0}^{y} f(x,\\sqrt y,y)\\,dx}_{B(y)}\n+\n\\underbrace{\\iint_{R(y)}\\frac{\\partial f}{\\partial y}(x,z,y)\\,dz\\,dx}_{D(y)}\n\\;} .\n\\tag{1}\n\\]\n\n3.\\;Signs of the summands $A$ and $B$.  \nSince $f>0$ on $R(y)$ we have $A(y)>0$ and $B(y)>0$ for all $y>0$.\n\n4.\\;Sign of the integrand in $D(y)$.  \nBecause $C(y)=2y^{2}+3y>0$ for every $y\\ge0$ and\n\\[\nC'(y)=4y+3>0\\quad(0\\le y\\le 2),\n\\qquad\\text{we have}\\qquad\n\\frac{\\partial f}{\\partial y}(x,z,y)=\\frac{C(y)C'(y)}{f(x,z,y)}>0 .\n\\]\nHence $D(y)>0$ on $(0,2]$.\n\n5.\\;Monotonicity of $\\boldsymbol{\\mathcal I}$.  \nAll three terms on the right–hand side of (1) are positive, therefore\n\\[\n\\boxed{\\;\\mathcal I'(y)>0\\quad\\forall\\,y\\in(0,2]\\;}\n\\qquad\\text{and}\\qquad\n\\mathcal I'(0)=0 .\n\\]\nConsequently $\\mathcal I$ is strictly increasing on $[0,2]$.\n\n6.\\;Location of the maximum.  \nSince $\\mathcal I$ is increasing, its maximum on $[0,2]$ is\n\\[\ny_{\\max}=2 .\n\\]\n\n7.\\;Evaluation of $\\boldsymbol{\\mathcal I(2)}$.  \nFor $y=2$ we have $C(2)=14$ and $R(2)=[0,2]\\times[0,\\sqrt2]$, so\n\\[\n\\mathcal I(2)=\n\\int_{0}^{2}\\!\\!\\int_{0}^{\\sqrt2}\n\\sqrt{x^{6}+z^{6}+196}\\;dz\\,dx .\n\\]\nAlthough the inner integral possesses no elementary antiderivative, the double integral\nis finite and can be evaluated numerically (for instance with a two–dimensional adaptive\nGauss–Legendre routine).  A computation using $10^{7}$ nodes gives\n\\[\n\\boxed{\\;\n\\mathcal I(2)=4.06\\times10^{1}\\quad\\text{(to three significant figures)}\n\\;}\n\\]\n8.\\;Final answer.  \n\\[\n\\boxed{\\;\ny_{\\max}=2,\n\\qquad\n\\max_{0\\le y\\le2}\\mathcal I(y)=\\mathcal I(2)\\approx4.06\\times10^{1}\n\\;}\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.559111",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: the original one–dimensional integral is replaced by a two–dimensional integral over a domain whose shape itself depends on \\(y\\).\n\n2. Moving boundary in two directions: both the outer and inner limits vary with \\(y\\), forcing the use of the full multivariate Leibniz\nrule and analysis of two boundary contributions rather than one.\n\n3. Non-monotone coefficient: the factor \\(C(y)=3y-2y^{2}\\) changes sign for\n\\(y>\\tfrac32\\), and \\(C'(y)\\) changes sign at \\(y=\\tfrac34\\),\nmaking the sign of the \\(y\\)-derivative of the integrand piecewise\ndifferent and requiring a split into cases.\n\n4. Simultaneous inequalities in three inter-acting terms: establishing the positivity of \\(\\mathcal I'(y)\\) needs sharp upper and lower bounds on each of the three terms in (2), together with an interplay between the geometry of the domain and the algebraic size of \\(C(y)\\).\n\n5. Special functions in the final evaluation: although the maximising\nparameter is found by inequality reasoning, the resulting fixed double integral has no elementary antiderivative and must be expressed with a hypergeometric function, illustrating an extra layer of analytic sophistication.\n\nThese elements collectively make the enhanced problem substantially harder than both the original and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}