Example 37 - If length of three sides of a trapezium other

Example 37 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 37 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 37 - Chapter 6 Class 12 Application of Derivatives - Part 4 Example 37 - Chapter 6 Class 12 Application of Derivatives - Part 5 Example 37 - Chapter 6 Class 12 Application of Derivatives - Part 6 Example 37 - Chapter 6 Class 12 Application of Derivatives - Part 7 Example 37 - Chapter 6 Class 12 Application of Derivatives - Part 8

  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Example 37 If length of three sides of a trapezium other than base are equal to 10 cm, then find the area of the trapezium when it is maximum.Let ABCD be given trapezium Given length of sides other than base is 10 โˆด AD = DC = CB = 10cm. Draw a perpendicular DP & CQ on AB. Let AP = ๐’™ cm By symmetry QB = ๐’™ cm Let A be the area of trapezium ABCD A = 1/2 (Sum of parallel sides) ร— (Height) A = ๐Ÿ/๐Ÿ (DC + AB) ร— DP Now, Since DP & CQ is perpendicular to AB, And CD was parallel to AB Thus, DPCQ forms a rectangle. โˆด PQ = DC = 10 cm Thus, AB = AP + PQ + QB = x + 10 + x = 2x + 10 Finding DP In ฮ” ADP By Pythagoras theorem DP2 + x2 = 102 DP2 + x2 = 100 DP2 = 100 โ€“ ๐‘ฅ2 DP = โˆš(๐Ÿ๐ŸŽ๐ŸŽ โˆ’๐’™๐Ÿ) From (1) A = ๐Ÿ/๐Ÿ (DC + AB) DP A = 1/2 (10+2๐‘ฅ+10) (โˆš(100โˆ’๐‘ฅ2)) A = 1/2 (2๐‘ฅ+20) (โˆš(100โˆ’๐‘ฅ2)) A = (2(๐‘ฅ +10) (โˆš(100 โˆ’ ๐‘ฅ2)))/2 A = (๐’™+๐Ÿ๐ŸŽ) โˆš(๐Ÿ๐ŸŽ๐ŸŽโˆ’๐’™๐Ÿ) We need to find area of trapezium when it is maximum i.e. We need to Maximize Area A = (๐‘ฅ+10) (โˆš(100โˆ’๐‘ฅ2)) Since A has a square root It will be difficult to differentiate Let Z = A2 = (๐‘ฅ+10)^2 (100โˆ’๐‘ฅ2) Since A is positive, A is maximum if A2 is maximum So, we maximize Z = A2 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 Using product rule As (๐‘ข๐‘ฃ)โ€ฒ = uโ€™v + vโ€™u 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’4(๐‘ฅ + 10)[๐‘ฅ^2+5๐‘ฅ+50] Putting ๐‘‘๐‘/๐‘‘๐‘ฅ=0 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‘ฅ + 10)(๐‘ฅโˆ’5)(๐‘ฅ+10)=0 (๐‘ฅ + 10)(๐‘ฅโˆ’5)(๐‘ฅ+10)=0 So, ๐‘ฅ=5 & ๐‘ฅ=โˆ’10 Since ๐‘ฅ represents distance & distance cannot be negative So, ๐’™ = 5 only By First Derivative Test Hence, ๐‘ฅ = 5 is point of Maxima โˆด Z is Maximum at ๐‘ฅ = 5 That means, Area A is maximum when 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 = 75โˆš3 Hence, maximum area is 75โˆš๐Ÿ‘ cm2

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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.