// Numbas version: exam_results_page_options {"name": "cormac's copy of 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": [{"preamble": {"js": "", "css": ""}, "parts": [{"scripts": {}, "showFeedbackIcon": true, "type": "gapfill", "showCorrectAnswer": true, "prompt": "

Input the value for \\(x\\) correct to one decimal place.

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\\(x = \\) [[0]]

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The length of the box is \\(\\var{l}-2x\\), the width is \\(\\var{w}-2x\\) and the height is \\(x\\).

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The volume is then given by

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\\(V=(\\var{l}-2x).(\\var{w}-2x).x\\)

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\\(V=\\simplify{4x^3-2*({w}+{l})x^2+{l}*{w}x}\\)

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\\(\\frac{dV}{dx}=\\simplify{12x^2-4*({w}+{l})x+{l}*{w}}\\)

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This is a quadratic equation.

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\\(x=\\frac{\\simplify{4*({w}+{l})}\\pm\\sqrt(\\simplify{16*({w}+{l})^2-48*{w}*{l}})}{24}\\)

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\\(x=\\frac{\\simplify{{w}+{l}}\\pm\\sqrt(\\simplify{{w}^2-{w}*{l}+{l}^2})}{6}\\)

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\\(\\frac{d^2V}{dx^2}=\\simplify{24x-4*({w}+{l})}\\)

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when  \\(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.

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Maximising the volume of a rectangular box

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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.

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Determine the value of \\(x\\) that will maximise the volume of this box.

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