Question 4 - Case Based Questions (MCQ) - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
An open box is to be made out of a piece of cardboard measuring (24 cm × 24 cm) by cutting of equal squares from the corners and turning up the sides.
Question 1
Find the volume of that open box?
(a) 4x
3
– 96x
2
+ 576x
(b) 4x
3
+ 96x
2
– 576x
(c) 2x
3
– 48x
2
+ 288x
(d) 2x
3
+ 48x
2
+ 288x
Question 2
Find the value of dV/dx
(a) 12(x
2
+ 16x – 48)
(b) 12(x
2
– 16x + 48)
(c) 6(x
2
+ 8x – 24)
(d) 6(x
2
– 8x + 24)
Question 3
Find the value of (d
2
V)/(dx
2
)
(a) 24(x + 8)
(b) 12(x – 4)
(c) 24(x – 8)
(d) 12(x + 4)
Question 4
Find the value of x other than 12?
(a) 3
(b) 9
(c) 1
(d) 4
Question 5
Volume is maximum at what height of that open box?
(a) 3 cm
(b) 9 cm
(c) 1 cm
(d) 4 cm
Transcript
Question An open box is to be made out of a piece of cardboard measuring (24 cm × 24 cm) by cutting of equal squares from the corners and turning up the sides. Based on the above information answer the following questions:
Question 1 Find the volume of that open box? (a) 4x3 – 96x2 + 576x (b) 4x3 + 96x2 – 576x (c) 2x3 – 48x2 + 288x (d) 2x3 + 48x2 + 288x
Here,
Height of box = x cm
Length of box = (24 − 2x) cm
Breadth of box = (24 − 2x) cm
Now,
Volume = Length × Breadth × Height
= (24 − 2x) × (24 − 2x) × x
= x(24 − 2x)2
= x(242 + 4x2 − 2 × 24 × 2x)
= x(576 + 4x2 − 96x)
= x(4x2 − 96x + 576)
= 4x3 − 96x2 + 576x
So, the correct answer is (A)
Question 2 Find the value of 𝑑𝑉/𝑑𝑥 (a) 12(x2 + 16x – 48) (b) 12(x2 – 16x + 48) (c) 6(x2 + 8x – 24) (d) 6(x2 – 8x + 24)
Now,
Volume V = 4x3 − 96x2 + 576x
Differentiating w.r.t. x
𝑑𝑉/𝑑𝑥=(𝑑(4𝑥^3 − 96𝑥^2 + 576𝑥))/𝑑𝑥
𝑑𝑉/𝑑𝑥=4 × 3𝑥^2−96 × 2𝑥+576
𝑑𝑉/𝑑𝑥=12𝑥^2−192𝑥+576
𝒅𝑽/𝒅𝒙=𝟏𝟐(𝒙^𝟐−𝟏𝟔𝒙+𝟒𝟖)
So, the correct answer is (B)
Question 3 Find the value of (𝑑^2 𝑉)/(𝑑𝑥^2 ) (a) 24(x + 8) (b) 12(x – 4) (c) 24(x – 8) (d) 12(x + 4)
Now,
𝑑𝑉/𝑑𝑥=12(𝑥^2−16𝑥+48)
Differentiating w.r.t x
(𝑑^2 𝑉)/(𝑑𝑥^2 )=𝑑(12(𝑥^2−16𝑥+48))/𝑑𝑥
(𝑑^2 𝑉)/(𝑑𝑥^2 )=12(2𝑥−16+0)
(𝑑^2 𝑉)/(𝑑𝑥^2 )=12(2𝑥−16)
(𝑑^2 𝑉)/(𝑑𝑥^2 )=12 × 2(𝑥−8)
(𝑑^2 𝑉)/(𝑑𝑥^2 )=𝟐𝟒(𝒙−𝟖)
So, the correct answer is (C)
Question 4 Find the value of x other than 12? (a) 3 (b) 9 (c) 1 (d) 4
Here we need to value of x for which Volume is Maximum
Putting 𝒅𝑽/𝒅𝒙=𝟎
12(𝑥^2−16𝑥+48)=0
𝑥^2−16𝑥+48=0
𝑥^2−12𝑥−4𝑥+48=0
𝑥(𝑥−12)−4(𝑥−12)=0
(𝑥−4) (𝑥−12)=0
So, x = 4, 12
Now, checking sign of
Volume V = 4x3 − 96x2 + 576x
= 4x(x2 − 24x + 144)
For x = 1
Volume = 4(1) (12 − 24(1) + 144)
So, the correct answer is (D)
Question 4 Volume is maximum at what height of that open box? (a) 3 cm (b) 9 cm (c) 1 cm (d) 4 cm
From Question 4
Volume ix maximum at x = 4 cm
So, the correct answer is (D)
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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