The area of a trapezium is defined by function 𝑓 and given by 𝑓(π‘₯) = (10 + π‘₯) √(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

Β 

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

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

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Question 34 The area of a trapezium is defined by function 𝑓 and given by 𝑓(π‘₯) = (10 + π‘₯) √("100 βˆ’ π‘₯2" ) , then the area when it is maximised is: (a) 75 cm2 (b) 7 √3 cm2 (c) 75 √3 cm2 (d) 5 cm2 𝑓(π‘₯) = (𝒙+𝟏𝟎) (√(πŸπŸŽπŸŽβˆ’π’™πŸ)) Since A has a square root It will be difficult to differentiate Let Z = [𝑓(π‘₯)]2 = (π‘₯+10)^2 (100βˆ’π‘₯2) Where f'(x) = 0, there Z’(x) = 0 Differentiating Z Z =(π‘₯+10)^2 " " (100βˆ’π‘₯2) Differentiating w.r.t. x Z’ = 𝑑((π‘₯ + 10)^2 " " (100 βˆ’ π‘₯2))/π‘‘π‘˜ Z’ = [(π‘₯ + 10)^2 ]^β€² (100 βˆ’ π‘₯^2 )+(π‘₯ + 10)^2 " " (100 βˆ’ π‘₯^2 )^β€² Z’ = 2(π‘₯ + 10)(100 βˆ’ π‘₯^2 )βˆ’2π‘₯(π‘₯ + 10)^2 Z’ = 2(π‘₯ + 10)[100 βˆ’ π‘₯^2βˆ’π‘₯(π‘₯+10)] Z’ = 2(π‘₯ + 10)[100 βˆ’ π‘₯^2βˆ’π‘₯^2βˆ’10π‘₯] Z’ = 2(π‘₯ + 10)[βˆ’2π‘₯^2βˆ’10π‘₯+100] Z’ = βˆ’πŸ’(𝒙 + 𝟏𝟎)[𝒙^𝟐+πŸ“π’™+πŸ“πŸŽ] Putting 𝒅𝒁/𝒅𝒙=𝟎 βˆ’4(π‘₯ + 10)[π‘₯^2+5π‘₯+50] =0 (π‘₯ + 10)[π‘₯^2+5π‘₯+50] =0 (π‘₯ + 10) [π‘₯2+10π‘₯βˆ’5π‘₯βˆ’50]=0 (π‘₯ + 10) [π‘₯(π‘₯+10)βˆ’5(π‘₯+10)]=0 (𝒙 + 𝟏𝟎)(π’™βˆ’πŸ“)(𝒙+𝟏𝟎)=𝟎 So, π‘₯=πŸ“ & 𝒙=βˆ’πŸπŸŽ Since x is length, it cannot be negative ∴ x = 5 Finding maximum area of trapezium A = (π‘₯+10) √(100βˆ’π‘₯2) = (5+10) √(100βˆ’(5)2) = (15) √(100βˆ’25) = 15 √75 = 15 √(25 Γ— 3) = 15 Γ— βˆšπŸπŸ“ Γ— βˆšπŸ‘ = 15 Γ— 5 Γ— √3 = 75βˆšπŸ‘ cm2 So, the correct answer is (C)

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

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.