Use Second Derivative Test to find the length 2x and width 2y of the soccer field (in terms of a and b) that maximize its area.
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CBSE Class 12 Sample Paper for 2023 Boards
CBSE Class 12 Sample Paper for 2023 Boards
Last updated at August 12, 2026 by Teachoo
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Transcript
Question 37 (iii) (Choice 2) Use Second Derivative Test to find the length 2x and width 2y of the soccer field (in terms of a and b) that maximize its area.Finding (š ^š š)/(š š^š ) šš/šš„=(32š^2)/š^2 Ć (š^2 š„ā2š„^3) Differentiating w.r.t š„ (š^2 š)/(šš„^2 )=(32š^2)/š^2 Ć (š^2ā2 Ć 3š„^2) (š ^š š)/(š š^š )=(ššš^š)/š^š Ć (š^šāšš^š) Putting x = š/āš (š^2 š)/(šš„^2 )=(32š^2)/š^2 Ć (š^2ā6(š/ā2)^2 ) (š^2 š)/(šš„^2 )=(32š^2)/š^2 Ć (š^2ā6 Ćš^2/2) (š^2 š)/(šš„^2 )=(32š^2)/š^2 Ć (š^2ā3š^2 ) (š^2 š)/(šš„^2 )=(32š^2)/š^2 Ć ā2š^2 < 0 Since š^ā²ā² < 0 for x = š/āš ā“ Z is maximum when x = š/āš Thus, A is maximum at x = š/āš Finding length 2x and 2y Length = 2x = 2 Ć š/ā2 = āša Breadth = 2y = 2 Ć (" " š)/š ā((š^š ā š^š ) ) = 2b/š Ćā(š^2ā(š/āš)^2 ) = 2b/š Ćā(š^2āš^2/2) = 2b/š Ćā(š^2/2) = 2b/š Ćš/āš = 2b/āš = āš š