An architect designs a building for a multi-national company. The floor consists of a rectangular region with semicircular ends having a perimeter of 200m as shown below:

An architect designs a building for a multi-national company..jpg

Based on the above information answer the following

Note: Check more Case Based Questions for - Applications of Derivatives Class 12 (Chapter 6 Class 12)

 

(i) If x and y represents the length and breadth of the rectangular region, then the relation between the variables is

a) x + π y = 100

b) 2x + π y = 200

c) π x + y = 50

d) x + y = 100

 

An architect designs a building for a multi-national company. The floo

Question 17 - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 17 - CBSE Class 12 Sample Paper for 2021 Boards - Part 3

 

(ii)The area of the rectangular region A expressed as a function of x is

a) 2 2/π (100 x – x 2 )

b) 1/π (100 x – x 2 )

c) x/π (100 – x)

d) πy 2 + 2/π (100x – x 2 )

 

Question 17 - CBSE Class 12 Sample Paper for 2021 Boards - Part 4 Question 17 - CBSE Class 12 Sample Paper for 2021 Boards - Part 5

 

(iii) The maximum value of area A is

a)  π/3200 m 2

b)  3200/π m 2

c)  5000/π m 2

d)  1000/π m 2

 

Question 17 - CBSE Class 12 Sample Paper for 2021 Boards - Part 6 Question 17 - CBSE Class 12 Sample Paper for 2021 Boards - Part 7

 

(iv) The CEO of the multi-national company is interested in maximizing the area of the whole floor including the semi-circular ends. For this to happen the valve of x should be

(a) 0 m

(b) 30 m

(c) 50 m

(d) 80 m

 

Question 17 - CBSE Class 12 Sample Paper for 2021 Boards - Part 8 Question 17 - CBSE Class 12 Sample Paper for 2021 Boards - Part 9 Question 17 - CBSE Class 12 Sample Paper for 2021 Boards - Part 10 Question 17 - CBSE Class 12 Sample Paper for 2021 Boards - Part 11

 

 

(v) The extra area generated if the area of the whole floor is maximized is

(a) 3000/π m 2

(b) 5000/π m 2

(c) 7000/π m 2

(d) No change Both areas are equal

 

Question 17 - CBSE Class 12 Sample Paper for 2021 Boards - Part 12 Question 17 - CBSE Class 12 Sample Paper for 2021 Boards - Part 13

Note: Check more Case Based Questions for - Applications of Derivatives Class 12 (Chapter 6 Class 12)

 

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Question 17 An architect designs a building for a multi-national company. The floor consists of a rectangular region with semicircular ends having a perimeter of 200m as shown below: Based on the above information answer the following Question 17 (i) If x and y represents the length and breadth of the rectangular region, then the relation between the variables is (a) x + π y = 100 (b) 2x + π y = 200 (c) π x + y = 50 (d) x + y = 100 Given that Perimeter of region = 200 m x + 1/2 × 2𝜋r + x + 1/2 × 2𝜋r = 200 2x + 2𝜋r = 200 Putting r = 𝑦/2 2x + 2𝜋 × 𝑦/2 = 200 2x + 𝜋y = 200 So, (b) is correct Question 17 (ii) The area of the rectangular region A expressed as a function of x is (a) 2/𝜋 (100 x – x2) (b) 1/𝜋 (100 x – x2) (c) 𝑥/𝜋 (100 – x) (d) 𝜋y2 + 2/𝜋 (100x – x2) Area A = Area of rectangle = x × y Since 2x + 𝜋y = 200 𝜋y = 200 − 2x y = 𝟏/𝝅 (200 − 2x) = x × 1/𝜋 (200 − 2x) = 1/𝜋 (200x − 2x2) = 𝟐/𝝅 (100x − x2) So, (a) is correct Question 17 (iii) The maximum value of area A is (a) 𝜋/3200 m2 (b) 3200/𝜋 m2 (c) 5000/𝜋 m2 (d) 1000/𝜋 m2 Now, A = 2/𝜋 (100x − x2) To find Maximum value of A We find 𝒅𝑨/𝒅𝒙 𝑑𝐴/𝑑𝑥=2/𝜋(100−2𝑥) Putting 𝒅𝑨/𝒅𝒙=𝟎 2/𝜋 (100−2𝑥)=0 100 = 2x x = 50 Putting x = 50 in value of A A = 2/𝜋 (100x − x2) A = 2/𝜋 (100 × 50 − (50)2) A = 2/𝜋 (5000 − 2500) A = 2/𝜋 × (2500) = 𝟓𝟎𝟎𝟎/(𝝅 ) m2 Question 17 (iv) The CEO of the multi-national company is interested in maximizing the area of the whole floor including the semi-circular ends. For this to happen the valve of x should be (a) 0 m (b) 30 m (c) 50 m (d) 80 m Total Area = A + 2 × Area of Semicircle = A + 2 × 1/2 × 𝜋r2 = A + 𝜋r2 Putting r = 𝑦/2 = A + 𝜋(𝑦/2)^2 = A + (𝜋y^2)/4 Putting value of A = 2/𝜋 (100x − x2) + (𝜋y^2)/4 Putting y = 𝟏/𝝅 (200 − 2x) = 2/𝜋 (100x − x2) + 𝜋/4 ×(1/𝜋 (200−2𝑥))^2 = 2/𝜋 (100x − x2) + 𝜋/4 ×1/𝜋^2 (200 − 2x)2 = 2/𝜋 (100x − x2) + 𝜋/4 ×1/𝜋^2 22 × (100 − x)2 = 2/𝜋 (100x − x2) + 1/𝜋 × (100 − x)2 = 2/𝜋 (100x − x2) + 1/𝜋 × (1002 + x2 − 200x) = 200𝑥/𝜋−(2𝑥^2)/𝜋+100^2/𝜋+𝑥^2/𝜋−200𝑥/(𝜋 ) = 200𝑥/𝜋−(2𝑥^2)/𝜋+100^2/𝜋+𝑥^2/𝜋−200𝑥/(𝜋 ) = (−𝑥^2)/𝜋+100^2/𝜋 Total Area = (−𝑥^2)/𝜋+100^2/𝜋 Let Total Area = Z Z = (−𝒙^𝟐)/𝝅+〖𝟏𝟎𝟎〗^𝟐/𝝅 To maximise Z, we find 𝑑𝑍/𝑑𝑥 𝑑𝑍/𝑑𝑥=𝑑((−𝑥^2)/𝜋 + 100^2/𝜋)/𝑑𝑥 𝑑𝑍/𝑑𝑥=−2𝑥/𝜋 Putting 𝒅𝒁/𝒅𝒙 = 0 −2𝑥/𝜋=0 𝒙 = 0 So, (a) is the correct answer Question 17 (v) The extra area generated if the area of the whole floor is maximized is (a) 3000/𝜋 m2 (b) 5000/𝜋 m2 (c) 7000/𝜋 m2 (d) No change Both areas are equal Total Area = Z = (−𝑥^2)/𝜋+100^2/𝜋 Putting x = 0 Z = (−0^2)/𝜋+100^2/𝜋 Z = 〖𝟏𝟎𝟎〗^𝟐/𝝅 Z = 𝟏𝟎𝟎𝟎𝟎/𝝅 m2 Also, Maximum Area A = 𝟓𝟎𝟎𝟎/(𝝅 ) m2 Thus, Difference between areas = 𝟏𝟎𝟎𝟎𝟎/𝝅 − 𝟓𝟎𝟎𝟎/(𝝅 ) = 𝟓𝟎𝟎𝟎/(𝝅 ) m2 So, correct answer is (b) Note: This answer doesn’t match the book, if we have done a mistake, please email – admin@teachoo.com

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.