Chapter 6 Class 12 Application of Derivatives
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Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Example 25 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 ā“ AD = DC = CB = 10cm. Draw a perpendicular DP & CQ on AB. Let AP = š cm By symmetry QB = š cm Let A be the area of trapezium ABCD A = 1/2 (Sum of parallel sides) Ć (Height) A = š/š (DC + AB) Ć DP Now, Since DP & CQ is perpendicular to AB, And CD was parallel to AB Thus, DPCQ forms a rectangle. ā“ PQ = DC = 10 cm Thus, AB = AP + PQ + QB = x + 10 + x = 2x + 10 Finding DP In Ī ADP By Pythagoras theorem DP2 + x2 = 102 DP2 + x2 = 100 DP2 = 100 ā š„2 DP = ā(ššš āšš) From (1) A = š/š (DC + AB) DP A = 1/2 (10+2š„+10) (ā(100āš„2)) A = 1/2 (2š„+20) (ā(100āš„2)) A = (2(š„ +10) (ā(100 ā š„2)))/2 A = (š+šš) ā(šššāšš) We need to find area of trapezium when it is maximum i.e. We need to Maximize Area A = (š„+10) (ā(100āš„2)) Since A has a square root It will be difficult to differentiate Let Z = A2 = (š„+10)^2 (100āš„2) Since A is positive, A is maximum if A2 is maximum So, we maximize Z = A2 Differentiating Z Z =(š„+10)^2 " " (100āš„2) Differentiating w.r.t. x Zā = š((š„ + 10)^2 " " (100 ā š„2))/šš Zā = [(š„ + 10)^2 ]^ā² (100 ā š„^2 )+(š„ + 10)^2 " " (100 ā š„^2 )^ā² Zā = 2(š„ + 10)(100 ā š„^2 )ā2š„(š„ + 10)^2 Zā = 2(š„ + 10)[100 ā š„^2āš„(š„+10)] Zā = 2(š„ + 10)[100 ā š„^2āš„^2ā10š„] Zā = 2(š„ + 10)[ā2š„^2ā10š„+100] Zā = ā4(š„ + 10)[š„^2+5š„+50] Putting šš/šš„=0 ā4(š„ + 10)[š„^2+5š„+50] =0 (š„ + 10)[š„^2+5š„+50] =0 (š„ + 10) [š„2+10š„ā5š„ā50]=0 (š„ + 10) [š„(š„+10)ā5(š„+10)]=0 (š„ + 10)(š„ā5)(š„+10)=0 (š„ + 10)(š„ā5)(š„+10)=0 So, š„=5 & š„=ā10 Since š„ represents distance & distance cannot be negative So, š = 5 only By First Derivative Test Hence, š„ = 5 is point of Maxima ā“ Z is Maximum at š„ = 5 That means, Area A is maximum when x = 5 Finding maximum area of trapezium A = (š„+10) ā(100āš„2) = (5+10) ā(100ā(5)2) = (15) ā(100ā25) = 15 ā75 = 75ā3 Hence, maximum area is 75āš cm2