Check sibling questions

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


Transcript

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)

  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo