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 April 16, 2024 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

(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

Made by

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

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