Question 17 (Case Based Question) - CBSE Class 12 Sample Paper for 2021 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Aug. 21, 2021 by Teachoo

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

(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

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(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
}
)

(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
}

(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

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

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

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.