Example 37 - Chapter 6 Class 12 Application of Derivatives

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Example 37 If length of three sides of a trapezium other than base are equal to 10cm, then find the area of the trapezium when it is maximum. Let ABCD be given trapezium & Given length of three side other than base is 10 ⇒ AD = DC = CB = 10cm. Draw a perpendicular DP & CQ on AB. Let AP = 𝑥 cm. By symmetry QB = 𝑥 cm. Let A be the area of trapezium ABCD A = 1/2 (Sum of parallel sides) × (Height) A = 1/2 (DC + AB) × DP Now, Since DP & CQ is perpendicular to AB, DPCQ forms a rectangle. ∴ PQ = DC = 10 cm Thus, AB = AP + PQ + QB = x + 10 + x = 2x + 10 & DC = 10 cm Finding DP In Δ ADP By Pythagoras theorem DP2 + x2 = 102 DP2 + x2 = 100 DP2 = 100 – 𝑥2 DP = √(100 −𝑥2) From (1) A = 1/2 (DC + AB) DP A = 1/2 (10+2𝑥+10) (√(100−𝑥2)) A = 1/2 (2𝑥+20) (√(100−𝑥2)) A = (2(𝑥 +10) (√(100 − 𝑥2)))/2 A = (𝑥+10) √(100−𝑥2) We need to find area of trapezium it is maximum i.e. we need to Maximize Area A = (𝑥+10) (√(100−𝑥2)) Diff w.r.t 𝑥 𝑑𝐴/𝑑𝑥= 𝑑((𝑥 + 10) (√(100 − 𝑥2)))/𝑑𝑥 𝑑𝐴/𝑑𝑥 = 𝑑(𝑥 − 10)/𝑑𝑥 . √(100−𝑥2) + 𝑑(√(100 − 𝑥2))/𝑑𝑥 . (𝑥+10) Using product rule As (𝑢𝑣)′ = u’v + v’u 𝑑𝐴/𝑑𝑥 = (1+0) √(100−𝑥2) + 1/(2√(100 − 𝑥^2 )) . 𝑑(100 − 𝑥2)/𝑑𝑥 . (𝑥+10) 𝑑𝐴/𝑑𝑥 = √(100−𝑥2) + 1/(2√(100 − 𝑥^2 )) . (0−2𝑥)(𝑥+10) 𝑑𝐴/𝑑𝑥 = √(100−𝑥2) – 2𝑥(𝑥 + 10)/(2√(100 − 𝑥^2 )) 𝑑𝐴/𝑑𝑥 = ((√(100 − 𝑥^2 ))^2− 𝑥(𝑥 + 10))/√(100 − 𝑥^2 ) 𝑑𝐴/𝑑𝑥 = ((100 − 𝑥^2 ) − 𝑥(𝑥 + 10))/((√(100 − 𝑥^2 )) ) 𝑑𝐴/𝑑𝑥 = (−2𝑥2− 10𝑥 + 100)/√(100 − 𝑥^2 ) Putting 𝑑𝐴/𝑑𝑥=0 (−2𝑥2− 10𝑥 + 100)/√(100 − 𝑥^2 )=0 −2𝑥2− 10𝑥 + 100=0 −2(𝑥^2+5𝑥−50)=0 𝑥2+5𝑥−50=0 𝑥2+10𝑥−5𝑥−50=0 𝑥(𝑥+10)−5(𝑥+10)=0 (𝑥−5)(𝑥+10)=0 So, 𝑥=5 & 𝑥=−10 Since 𝑥 represents distance & distance cannot be negative So, 𝑥 = 5 only. Hence value of 𝑥 = 5 Hence, 𝑥 = 5 is point of Maxima ∴ A is Maximum at 𝑥 = 5 Finding maximum area of trapezium A = (𝑥+10) √(100−𝑥2) = (5+10) √(100−(5)2) = (15) √(100−25) = 15 √75 = 75√3 Hence, maximum area is 75√𝟑 cm2