// Numbas version: finer_feedback_settings {"name": "cormac's copy of Maximum/minimum", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Untitled.jpg", "/srv/numbas/media/question-resources/Untitled.jpg"], ["question-resources/Untitled_qCawkyB.jpg", "/srv/numbas/media/question-resources/Untitled_qCawkyB.jpg"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"rulesets": {}, "ungrouped_variables": ["l", "w"], "variables": {"w": {"description": "", "name": "w", "definition": "random(10..28#2)", "templateType": "randrange", "group": "Ungrouped variables"}, "l": {"description": "", "name": "l", "definition": "random(30..100#5)", "templateType": "randrange", "group": "Ungrouped variables"}}, "metadata": {"description": "
Maximising the volume of a rectangular box
", "licence": "Creative Commons Attribution 4.0 International"}, "variable_groups": [], "tags": [], "extensions": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "parts": [{"prompt": "Input the value for \\(x\\) correct to one decimal place.
\n\\(x = \\) [[0]]
", "gaps": [{"scripts": {}, "showCorrectAnswer": true, "showPrecisionHint": false, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "type": "numberentry", "minValue": "({w}+{l}-sqrt({w}^2+{l}^2-{w}*{l}))/6", "correctAnswerFraction": false, "precision": "1", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "maxValue": "({w}+{l}-sqrt({w}^2+{l}^2-{w}*{l}))/6", "marks": 1, "strictPrecision": false, "precisionType": "dp", "allowFractions": false}], "scripts": {}, "variableReplacementStrategy": "originalfirst", "marks": 0, "variableReplacements": [], "type": "gapfill", "showCorrectAnswer": true}], "preamble": {"js": "", "css": ""}, "functions": {}, "name": "cormac's copy of Maximum/minimum", "statement": "A rectangular sheet of metal of length = \\(\\var{l}cm\\) and width = \\(\\var{w}cm\\) has a square of side \\(x\\,cm\\) cut from each corner. The ends and sides will be folded upwards to form an open box.
\nDetermine the value of \\(x\\) that will maximise the volume of this box.
\n", "advice": "The length of the box is \\(\\var{l}-2x\\), the width is \\(\\var{w}-2x\\) and the height is \\(x\\).
\nThe volume is then given by
\n\\(V=(\\var{l}-2x).(\\var{w}-2x).x\\)
\n\\(V=\\simplify{4x^3-2*({w}+{l})x^2+{l}*{w}x}\\)
\n\\(\\frac{dV}{dx}=\\simplify{12x^2-4*({w}+{l})x+{l}*{w}}\\)
\nThis is a quadratic equation.
\n\\(x=\\frac{\\simplify{4*({w}+{l})}\\pm\\sqrt(\\simplify{16*({w}+{l})^2-48*{w}*{l}})}{24}\\)
\n\\(x=\\frac{\\simplify{{w}+{l}}\\pm\\sqrt(\\simplify{{w}^2-{w}*{l}+{l}^2})}{6}\\)
\n\\(\\frac{d^2V}{dx^2}=\\simplify{24x-4*({w}+{l})}\\)
\nwhen \\(x=\\simplify{({w}+{l}-sqrt({w}^2-{w}*{l}+{l}^2))/6}\\) \\(\\frac{d^2V}{dx^2}<0\\) and therefore is the value that gives a maximum.
", "type": "question", "contributors": [{"name": "cormac breen", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/306/"}]}]}], "contributors": [{"name": "cormac breen", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/306/"}]}