Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month
Example 37 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at April 19, 2021 by Teachoo
Examples
Example 1
Example 2
Example 3
Example 4 Important
Example 5
Example 6
Example 7
Example 8 Important
Example 9 Important
Example 10
Example 11 Important
Example 12
Example 13 Important
Example 14
Example 15
Example 16
Example 17 Important
Example 18
Example 19
Example 20
Example 21
Example 22
Example 23
Example 24
Example 25
Example 26
Example 27
Example 28 Important
Example 29
Example 30 Important
Example 31
Example 32 Important
Example 33 Important
Example 34
Example 35 Important
Example 36
Example 37 Important You are here
Example 38 Important
Example 39
Example 40 Important
Example 41 Important
Example 42 Important
Example 43 Important
Example 44 Important
Example 45 Important
Example 46 Important
Example 47 Important
Example 48 Important
Example 49
Example 50 Important
Example 51
Chapter 6 Class 12 Application of Derivatives
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 β΄ AD = DC = CB = 10cm. Draw a perpendicular DP & CQ on AB. Let AP = π cm By symmetry QB = π cm Let A be the area of trapezium ABCD A = 1/2 (Sum of parallel sides) Γ (Height) A = π/π (DC + AB) Γ DP Now, Since DP & CQ is perpendicular to AB, And CD was parallel to AB Thus, DPCQ forms a rectangle. β΄ PQ = DC = 10 cm Thus, AB = AP + PQ + QB = x + 10 + x = 2x + 10 Finding DP In Ξ ADP By Pythagoras theorem DP2 + x2 = 102 DP2 + x2 = 100 DP2 = 100 β π₯2 DP = β(πππ βππ) From (1) A = π/π (DC + AB) DP A = 1/2 (10+2π₯+10) (β(100βπ₯2)) A = 1/2 (2π₯+20) (β(100βπ₯2)) A = (2(π₯ +10) (β(100 β π₯2)))/2 A = (π+ππ) β(πππβππ) We need to find area of trapezium when it is maximum i.e. We need to Maximize Area A = (π₯+10) (β(100βπ₯2)) Since A has a square root It will be difficult to differentiate Let Z = A2 = (π₯+10)^2 (100βπ₯2) Since A is positive, A is maximum if A2 is maximum So, we maximize Z = A2 Differentiating Z Z =(π₯+10)^2 " " (100βπ₯2) Differentiating w.r.t. x Zβ = π((π₯ + 10)^2 " " (100 β π₯2))/ππ Zβ = [(π₯ + 10)^2 ]^β² (100 β π₯^2 )+(π₯ + 10)^2 " " (100 β π₯^2 )^β² Zβ = 2(π₯ + 10)(100 β π₯^2 )β2π₯(π₯ + 10)^2 Using product rule As (π’π£)β² = uβv + vβu Zβ = 2(π₯ + 10)[100 β π₯^2βπ₯(π₯+10)] Zβ = 2(π₯ + 10)[100 β π₯^2βπ₯^2β10π₯] Zβ = 2(π₯ + 10)[β2π₯^2β10π₯+100] Zβ = β4(π₯ + 10)[π₯^2+5π₯+50] Putting ππ/ππ₯=0 β4(π₯ + 10)[π₯^2+5π₯+50] =0 (π₯ + 10)[π₯^2+5π₯+50] =0 (π₯ + 10) [π₯2+10π₯β5π₯β50]=0 (π₯ + 10) [π₯(π₯+10)β5(π₯+10)]=0 (π₯ + 10)(π₯β5)(π₯+10)=0 (π₯ + 10)(π₯β5)(π₯+10)=0 So, π₯=5 & π₯=β10 Since π₯ represents distance & distance cannot be negative So, π = 5 only By First Derivative Test Hence, π₯ = 5 is point of Maxima β΄ Z is Maximum at π₯ = 5 That means, Area A is maximum when x = 5 Finding maximum area of trapezium A = (π₯+10) β(100βπ₯2) = (5+10) β(100β(5)2) = (15) β(100β25) = 15 β75 = 75β3 Hence, maximum area is 75βπ cm2
