The area of a trapezium is defined by function š‘“ and given by š‘“(š‘„) = (10 + š‘„) √(100 - x 2 ) , then the area when it is maximised is:

(a) 75 cm 2          (b) 7 √3 cm 2

(c) 75 √3 cm 2     (d) 5 cm 2

 

This question is inspired from Example 37 - Chapter 6 Class 12   - Application of Derivatives

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Question 34 The area of a trapezium is defined by function š‘“ and given by š‘“(š‘„) = (10 + š‘„) āˆš("100 āˆ’ š‘„2" ) , then the area when it is maximised is: (a) 75 cm2 (b) 7 āˆš3 cm2 (c) 75 āˆš3 cm2 (d) 5 cm2 š‘“(š‘„) = (š’™+šŸšŸŽ) (āˆš(šŸšŸŽšŸŽāˆ’š’™šŸ)) Since A has a square root It will be difficult to differentiate Let Z = [š‘“(š‘„)]2 = (š‘„+10)^2 (100āˆ’š‘„2) Where f'(x) = 0, there Zā€™(x) = 0 Differentiating Z Z =(š‘„+10)^2 " " (100āˆ’š‘„2) Differentiating w.r.t. x Zā€™ = š‘‘((š‘„ + 10)^2 " " (100 āˆ’ š‘„2))/š‘‘š‘˜ Zā€™ = [(š‘„ + 10)^2 ]^ā€² (100 āˆ’ š‘„^2 )+(š‘„ + 10)^2 " " (100 āˆ’ š‘„^2 )^ā€² Zā€™ = 2(š‘„ + 10)(100 āˆ’ š‘„^2 )āˆ’2š‘„(š‘„ + 10)^2 Zā€™ = 2(š‘„ + 10)[100 āˆ’ š‘„^2āˆ’š‘„(š‘„+10)] Zā€™ = 2(š‘„ + 10)[100 āˆ’ š‘„^2āˆ’š‘„^2āˆ’10š‘„] Zā€™ = 2(š‘„ + 10)[āˆ’2š‘„^2āˆ’10š‘„+100] Zā€™ = āˆ’šŸ’(š’™ + šŸšŸŽ)[š’™^šŸ+šŸ“š’™+šŸ“šŸŽ] Putting š’…š’/š’…š’™=šŸŽ āˆ’4(š‘„ + 10)[š‘„^2+5š‘„+50] =0 (š‘„ + 10)[š‘„^2+5š‘„+50] =0 (š‘„ + 10) [š‘„2+10š‘„āˆ’5š‘„āˆ’50]=0 (š‘„ + 10) [š‘„(š‘„+10)āˆ’5(š‘„+10)]=0 (š’™ + šŸšŸŽ)(š’™āˆ’šŸ“)(š’™+šŸšŸŽ)=šŸŽ So, š‘„=šŸ“ & š’™=āˆ’šŸšŸŽ Since x is length, it cannot be negative āˆ“ x = 5 Finding maximum area of trapezium A = (š‘„+10) āˆš(100āˆ’š‘„2) = (5+10) āˆš(100āˆ’(5)2) = (15) āˆš(100āˆ’25) = 15 āˆš75 = 15 āˆš(25 Ɨ 3) = 15 Ɨ āˆššŸšŸ“ Ɨ āˆššŸ‘ = 15 Ɨ 5 Ɨ āˆš3 = 75āˆššŸ‘ cm2 So, the correct answer is (C)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.