The area of a trapezium is defined by function ๐‘“ and given by ๐‘“(๐‘ฅ) = (10 + ๐‘ฅ) √(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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  1. Class 12
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Transcript

Question 34 The area of a trapezium is defined by function ๐‘“ and given by ๐‘“(๐‘ฅ) = (10 + ๐‘ฅ) โˆš("100 โˆ’ ๐‘ฅ2" ) , then the area when it is maximised is: (a) 75 cm2 (b) 7 โˆš3 cm2 (c) 75 โˆš3 cm2 (d) 5 cm2 ๐‘“(๐‘ฅ) = (๐’™+๐Ÿ๐ŸŽ) (โˆš(๐Ÿ๐ŸŽ๐ŸŽโˆ’๐’™๐Ÿ)) Since A has a square root It will be difficult to differentiate Let Z = [๐‘“(๐‘ฅ)]2 = (๐‘ฅ+10)^2 (100โˆ’๐‘ฅ2) Where f'(x) = 0, there Zโ€™(x) = 0 Differentiating Z Z =(๐‘ฅ+10)^2 " " (100โˆ’๐‘ฅ2) Differentiating w.r.t. x Zโ€™ = ๐‘‘((๐‘ฅ + 10)^2 " " (100 โˆ’ ๐‘ฅ2))/๐‘‘๐‘˜ Zโ€™ = [(๐‘ฅ + 10)^2 ]^โ€ฒ (100 โˆ’ ๐‘ฅ^2 )+(๐‘ฅ + 10)^2 " " (100 โˆ’ ๐‘ฅ^2 )^โ€ฒ Zโ€™ = 2(๐‘ฅ + 10)(100 โˆ’ ๐‘ฅ^2 )โˆ’2๐‘ฅ(๐‘ฅ + 10)^2 Zโ€™ = 2(๐‘ฅ + 10)[100 โˆ’ ๐‘ฅ^2โˆ’๐‘ฅ(๐‘ฅ+10)] Zโ€™ = 2(๐‘ฅ + 10)[100 โˆ’ ๐‘ฅ^2โˆ’๐‘ฅ^2โˆ’10๐‘ฅ] Zโ€™ = 2(๐‘ฅ + 10)[โˆ’2๐‘ฅ^2โˆ’10๐‘ฅ+100] Zโ€™ = โˆ’๐Ÿ’(๐’™ + ๐Ÿ๐ŸŽ)[๐’™^๐Ÿ+๐Ÿ“๐’™+๐Ÿ“๐ŸŽ] Putting ๐’…๐’/๐’…๐’™=๐ŸŽ โˆ’4(๐‘ฅ + 10)[๐‘ฅ^2+5๐‘ฅ+50] =0 (๐‘ฅ + 10)[๐‘ฅ^2+5๐‘ฅ+50] =0 (๐‘ฅ + 10) [๐‘ฅ2+10๐‘ฅโˆ’5๐‘ฅโˆ’50]=0 (๐‘ฅ + 10) [๐‘ฅ(๐‘ฅ+10)โˆ’5(๐‘ฅ+10)]=0 (๐’™ + ๐Ÿ๐ŸŽ)(๐’™โˆ’๐Ÿ“)(๐’™+๐Ÿ๐ŸŽ)=๐ŸŽ So, ๐‘ฅ=๐Ÿ“ & ๐’™=โˆ’๐Ÿ๐ŸŽ Since x is length, it cannot be negative โˆด x = 5 Finding maximum area of trapezium A = (๐‘ฅ+10) โˆš(100โˆ’๐‘ฅ2) = (5+10) โˆš(100โˆ’(5)2) = (15) โˆš(100โˆ’25) = 15 โˆš75 = 15 โˆš(25 ร— 3) = 15 ร— โˆš๐Ÿ๐Ÿ“ ร— โˆš๐Ÿ‘ = 15 ร— 5 ร— โˆš3 = 75โˆš๐Ÿ‘ cm2 So, the correct answer is (C)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.