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Last updated at April 19, 2021 by Teachoo

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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 โด 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 = ๐/๐ (DC + AB) ร DP Now, Since DP & CQ is perpendicular to AB, And CD was parallel to AB Thus, DPCQ forms a rectangle. โด 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 ๐ฅ2 DP = โ(๐๐๐ โ๐๐) From (1) A = ๐/๐ (DC + AB) DP A = 1/2 (10+2๐ฅ+10) (โ(100โ๐ฅ2)) A = 1/2 (2๐ฅ+20) (โ(100โ๐ฅ2)) A = (2(๐ฅ +10) (โ(100 โ ๐ฅ2)))/2 A = (๐+๐๐) โ(๐๐๐โ๐๐) We need to find area of trapezium when it is maximum i.e. We need to Maximize Area A = (๐ฅ+10) (โ(100โ๐ฅ2)) Since A has a square root It will be difficult to differentiate Let Z = A2 = (๐ฅ+10)^2 (100โ๐ฅ2) Since A is positive, A is maximum if A2 is maximum So, we maximize Z = A2 Differentiating Z Z =(๐ฅ+10)^2 " " (100โ๐ฅ2) Differentiating w.r.t. x Zโ = ๐((๐ฅ + 10)^2 " " (100 โ ๐ฅ2))/๐๐ Zโ = [(๐ฅ + 10)^2 ]^โฒ (100 โ ๐ฅ^2 )+(๐ฅ + 10)^2 " " (100 โ ๐ฅ^2 )^โฒ Zโ = 2(๐ฅ + 10)(100 โ ๐ฅ^2 )โ2๐ฅ(๐ฅ + 10)^2 Using product rule As (๐ข๐ฃ)โฒ = uโv + vโu Zโ = 2(๐ฅ + 10)[100 โ ๐ฅ^2โ๐ฅ(๐ฅ+10)] Zโ = 2(๐ฅ + 10)[100 โ ๐ฅ^2โ๐ฅ^2โ10๐ฅ] Zโ = 2(๐ฅ + 10)[โ2๐ฅ^2โ10๐ฅ+100] Zโ = โ4(๐ฅ + 10)[๐ฅ^2+5๐ฅ+50] Putting ๐๐/๐๐ฅ=0 โ4(๐ฅ + 10)[๐ฅ^2+5๐ฅ+50] =0 (๐ฅ + 10)[๐ฅ^2+5๐ฅ+50] =0 (๐ฅ + 10) [๐ฅ2+10๐ฅโ5๐ฅโ50]=0 (๐ฅ + 10) [๐ฅ(๐ฅ+10)โ5(๐ฅ+10)]=0 (๐ฅ + 10)(๐ฅโ5)(๐ฅ+10)=0 (๐ฅ + 10)(๐ฅโ5)(๐ฅ+10)=0 So, ๐ฅ=5 & ๐ฅ=โ10 Since ๐ฅ represents distance & distance cannot be negative So, ๐ = 5 only By First Derivative Test Hence, ๐ฅ = 5 is point of Maxima โด Z is Maximum at ๐ฅ = 5 That means, Area A is maximum when x = 5 Finding maximum area of trapezium A = (๐ฅ+10) โ(100โ๐ฅ2) = (5+10) โ(100โ(5)2) = (15) โ(100โ25) = 15 โ75 = 75โ3 Hence, maximum area is 75โ๐ cm2

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Chapter 6 Class 12 Application of Derivatives (Term 1)

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About the Author

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.