The area of a trapezium is defined by function 𝑓 and given by 𝑓(𝑥) = (10 + 𝑥) √(100 - x 2 ) , then the area when it is maximised is:

(a) 75 cm 2          (b) 7 √3 cm 2

(c) 75 √3 cm 2     (d) 5 cm 2

 

This question is inspired from Example 37 - Chapter 6 Class 12   - Application of Derivatives

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Question 34 The area of a trapezium is defined by function 𝑓 and given by 𝑓(𝑥) = (10 + 𝑥) √("100 − 𝑥2" ) , then the area when it is maximised is: (a) 75 cm2 (b) 7 √3 cm2 (c) 75 √3 cm2 (d) 5 cm2 𝑓(𝑥) = (𝒙+𝟏𝟎) (√(𝟏𝟎𝟎−𝒙𝟐)) Since A has a square root It will be difficult to differentiate Let Z = [𝑓(𝑥)]2 = (𝑥+10)^2 (100−𝑥2) Where f'(x) = 0, there Z’(x) = 0 Differentiating Z Z =(𝑥+10)^2 " " (100−𝑥2) Differentiating w.r.t. x Z’ = 𝑑((𝑥 + 10)^2 " " (100 − 𝑥2))/𝑑𝑘 Z’ = [(𝑥 + 10)^2 ]^′ (100 − 𝑥^2 )+(𝑥 + 10)^2 " " (100 − 𝑥^2 )^′ Z’ = 2(𝑥 + 10)(100 − 𝑥^2 )−2𝑥(𝑥 + 10)^2 Z’ = 2(𝑥 + 10)[100 − 𝑥^2−𝑥(𝑥+10)] Z’ = 2(𝑥 + 10)[100 − 𝑥^2−𝑥^2−10𝑥] Z’ = 2(𝑥 + 10)[−2𝑥^2−10𝑥+100] Z’ = −𝟒(𝒙 + 𝟏𝟎)[𝒙^𝟐+𝟓𝒙+𝟓𝟎] Putting 𝒅𝒁/𝒅𝒙=𝟎 −4(𝑥 + 10)[𝑥^2+5𝑥+50] =0 (𝑥 + 10)[𝑥^2+5𝑥+50] =0 (𝑥 + 10) [𝑥2+10𝑥−5𝑥−50]=0 (𝑥 + 10) [𝑥(𝑥+10)−5(𝑥+10)]=0 (𝒙 + 𝟏𝟎)(𝒙−𝟓)(𝒙+𝟏𝟎)=𝟎 So, 𝑥=𝟓 & 𝒙=−𝟏𝟎 Since x is length, it cannot be negative ∴ x = 5 Finding maximum area of trapezium A = (𝑥+10) √(100−𝑥2) = (5+10) √(100−(5)2) = (15) √(100−25) = 15 √75 = 15 √(25 × 3) = 15 × √𝟐𝟓 × √𝟑 = 15 × 5 × √3 = 75√𝟑 cm2 So, the correct answer is (C)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.