case study questions on application of derivatives

Class 12 Maths: Case Study of Chapter 6 Applications of Derivatives PDF Download

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In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Mathematics Chapter 6 Applications of Derivatives Case Study and Passage Based Questions with Answers were Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Maths Applications of Derivatives  to know their preparation level.

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In CBSE Class 12 Maths Paper, There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Applications of Derivatives Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 12 Mathematics  Chapter 6 Applications of Derivatives

Case Study/Passage-Based Questions

(i) To construct a garden using 200 ft of fencing, we need to maximize its

Answer: (b) area

(ii) If x denote the length of side of garden perpendicular to brick wall and y denote the length, of side parallel to brick wall, then find the relation representing total amount of fencing wire.

Answer: (c) y+2x=200

(iii) Area of the garden as a function of x, say A(x), can be represented as

Answer: (c) 200x – 2×2

(iv) Maximum value of A(x) occurs at x equals

Answer: (a) 50 ft

(v) Maximum area of garden will be

Answer: (c) 5000sq.ft

Hope the information shed above regarding Case Study and Passage Based Questions for Class 12 Maths Chapter 6 Applications of Derivatives with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 12 Mathematics Applications of Derivatives Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

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CBSE Case Study Questions for Class 12 Maths Applications of Derivatives Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions for Class 12 Maths Applications of Derivatives  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams! I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 12 Maths Applications of Derivatives PDF

Checkout our case study questions for other chapters.

• Chapter 4 Determinants Case Study Questions
• Chapter 5 Continuity and Differentiability Case Study Questions
• Chapter 7 Integrals Case Study Questions
• Chapter 8 Applications of Integrals Case Study Questions

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Class 12th Maths - Application of Derivatives Case Study Questions and Answers 2022 - 2023

Class 12th Maths - Application of Derivatives Case Study Questions and Answers 2022 - 2023 Study Materials Sep-08 , 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 12 Maths Subject - Application of Derivatives, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Application of derivatives case study questions with answer key.

Final Semester - June 2015

(ii) Volume of the open box formed by folding up the cutting corner can be expressed as

(iii) The values of x for which  \(\begin{equation} \frac{d V}{d x}=0 \end{equation}\)  ,are

(iv) Megha is interested in maximising the volume of the box. So, what should be the side of the square to be cut off so that the volume of the box is maximum?

(v) The maximum value of the volume is

(ii) If x denote the length of side of garden perpendicular to brick wall and y denote the length, of side parallel to brick wall, then find the relation representing total amount of fencing wire.

(iii) Area of the garden as a function of x, say A(x), can be represented as

(iv) Maximum value of A(x) occurs at x equals

(v) Maximum area of garden will be

(ii) Revenue R as a function of x can be represented as

(iii) Find the number of days after 1stJuly, when Shyams father attain maximum revenue.

(iv) On which day should Shyam's father harvest the onions to maximise his revenue?

(v) Maximum revenue is equal to

An owner of an electric bi~e rental company have determined that if they charge customers Rs. x per day to rent a bike, where 50 Rs. x Rs. 200, then number of bikes (n), they rent per day can be shown by linear function n(x) = 2000 - 10x. If they charge Rs. 50 per day or less, they will rent all their bikes. If they charge Rs. 200 or more per day, they will not rent any bike. Based on the above information, answer the following questions.

(ii) If R(x) denote the revenue, then maximum value of R(x) occur when x equals

(iii) At x = 260, the revenue collected by the company is

(iv) The number of bikes rented per day, if x = 105 is

(v) Maximum revenue collected by company is

(ii) If x represent the number of apartments which are not rented, then the profit expressed as a function of x is

(iii) If P = 10500, then N =

(iv) If P = 11,000, then the profit is

(ii) The range of x is

(iii) The value of xfor which revenue is maximum, is

(iv) When the revenue is maximum, the price of the ticket is

(v) How any spectators should be present to maximize the revenue?

(ii) The radius that will minimize the cost of the material to manufacture the tin can is

(iii) The height thatt will minimize the cost of the material to manufacture the tin can is

(iv) If the cost of material used to manufacture the tin can is Rs.100/m 2 and  \(\sqrt[3]{\frac{1500}{\pi}} \approx 7.8\)  then minimum cost is approximately

(v) To minimize the cost of the material used to manufacture the tin can, we need to minimize the

(ii) The relation between a and b is given by

(iii) Area of poster in terms of b is given by

(iv) The value of b, so that area of the poster is minimized, is

(v) The value of a, so that area of the poster is minimized, is

(i) In order to make a least expensive water tank, Nitin need to minimize its

(ii) Total cost of tank as a function of h can' be' represented as

(iii) Range of h is

(iv) Value of h at which c(h) is minimum, is

(v) The cost ofleast expensive tank is

(ii) If sum of the surface areas of box and ball are given to be constant k 2 , then x is equal to

(iii) The radius of the ball, when S is minimum, is

(iv) Relation between length of the box and radius of the ball can be represented as

(v) Minimum value of S is

(ii) If C(x) denote the maintenance cost function, then maximum value of C(x) occur at x =

(iii) The maximum value of C(x) would be

(iv) The number of apartments, that the complex should have in order to minimize the maintenance cost, is

(v) If the minimum maintenance cost is attain, then the maintenance cost for each apartment would be

(ii) If x is the length of the outer rectangle, then area of inner rectangle in terms of x is

(iii) Find the range of x.

(iv) If area of inner rectangle is m~imum, then x is equal to

(v) If area of inner rectangle is maximum, then length and breadth of this rectangle are respectively

(ii) If magazine company increases Rs.500 as annual charges, then R is equal to

(iv) What amount of increase in annual charges will bring maximum revenue?

(ii) The maximum value of x can not be

(iii) The rainimum value of x can not be

(iv) If l(x) denote the combined light intensity, then lex) will be minimum when x =

(v) The darkest spot between the two lights is

(ii) The relation between x and y is

(iii) The outer surface area of tank will be minimum when depth of tank is equal to

(iv) The cost of material will be least when width of tank is equal to

(v) If cost of aluminium sheet is Rs.360/m 2 , then the minimum cost for the construction of tank will be

(ii) Distance (say D) between Arun and Manita will be

(iii) For which real value(s) of x, first derivative of D 2 w.r.t. 'x' will  Vanish?

(iv) Find the position of Arun when Manita will hit the paper hall.

(v) The minimum value of D is

(ii) The area (A) of green grass, in terms of x, is given by

(iii) The maximum value of A is

(iv) The value oflength of rectangle, when A is maximum, is

(v) The area of gravelling path is

(ii) The length PQ is

(iii) Let there be a quantity S such that S = Rp 2 + RQ 2 , then S is given by

(iv) Find the value of x for which value of S is minimum.

(v) For minimum value of S, find the value of PR and RQ

(ii) The area (A) of the window can be given by

(iii) Rohan is interested in maximizing the area of the whole window, for this to happen, the value of x should be

(iv) Maximum area of the window is

(v) For maximum value of A, the breadth of rectangular part of the window is

(ii) The area (A) of the rectangular region, as a function of x, can be expressed as

(iii) School's manager is interested in maximising the area of floor 'A' for this to be happen, the value of x should be

(iv) The value of y, for which the area of floor is maximum is

(v) Maximum area of floor is

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4: Applications of Derivatives

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A rocket launch involves two related quantities that change over time. Being able to solve this type of problem is just one application of derivatives introduced in this chapter. We also look at how derivatives are used to find maximum and minimum values of functions. As a result, we will be able to solve applied optimization problems, such as maximizing revenue and minimizing surface area. In addition, we examine how derivatives are used to evaluate complicated limits, to approximate roots of functions, and to provide accurate graphs of functions.

• 4.1: Prelude to Applications of Derivatives A rocket launch involves two related quantities that change over time. Being able to solve this type of problem is just one application of derivatives introduced in this chapter. We also look at how derivatives are used to find maximum and minimum values of functions. As a result, we will be able to solve applied optimization problems, such as maximizing revenue and minimizing surface area. In addition, we examine how derivatives are used to evaluate complicated limits, to approximate roots of f
• 4.2E: Exercises for Section 4.1
• 4.3E: Exercises for Section 4.2
• 4.4E: Exercises for Section 4.3
• 4.5E: Exercises for Section 4.4
• 4.6E: Exercises for Section 4.5
• 4.7E: Exercises for Section 4.6
• 4.8E: Exercises for Section 4.7
• 4.9E: Exercises for Section 4.8
• 4.10E: Exercises for Section 4.9
• 4.11E: Exercises for Section 4.10
• 4.12: Chapter 4 Review Exercises

Contributors and Attributions

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .

Class 12 CBSE A.O.D ML AGGARWAL Case Study

Please select, case-study-1 a political party placed an order to a screen printer for the printing of party's slogan on a rectangular cloth sheets. a margin of 2.5 cm along length and 1 cm along width of cloth was left. the pictorial view of the slogan in shown below: if the total area of cloth is 640 cm2, based on the above information, answer the following questions: (i) the length and the breadth (in cm) of the rectangular part on which the slogan is printed respectively are (a)x-2.5, y-1 (b) x-5,y-2 (c) x-2,y-5 (d) x-5,y-1 (ii) the relation between x and y is (a) (x-2.5)(y-1)=640 (b) (x-5)(y-2)=640 (c) xy=640 (d) (x-2)(y-5)= (iii) area 'a' of cloth on which slogan matter was written is given by (a) 650-2x-3200/x (b) 650+2x+ 3200/x (c) 600-2x-3200/x (d) 600 +2x + 3200/x (iv) what is the value of x for which the printing area is maximum (a) 16 cm (b) 40 cm (c) 14 cm (d) 11 cm (v) what is the value of 'a' when printing of slogan is done on maximum area (a) 640 cm² (b) 418 cm² (c) 490 cm² (d) none of these.

Case-study-2 Saloni has a piece of tin rectangular in shape as shown in the diagram given below. She is going to cut squares from each corners and fold up the sides to form an open box. Based on the above information, answer the following questions: (i) If x is the side of square cut off from each corner of the sheet, then what is the length of open box? (a) 45-x (b) 45-2x (c) 45 +2x (d) none of these (ii) What is width of open box? (a) 24-x (b) 24+ x (c) 24-2x (d) none of these (iii) Volume 'V' of open box is given by (a) V = (45-x)(24-x) x (b) V=(45+x)(24+x) x (c) V=(45+2x) (24+2x) x (d) V=(45-2x)(24-2x) x (iv) What should be the side of square to be cut off so that the volume of box is maximum? (a) 18 cm (b) 5 cm (c) both (a) and (b) (d) none of these (v) The maximum value of V (in cm³) is (a) 5400 (b) 4200 (c) 3600 (d) 2450

Case-study-3 A company is planning to launch a new product and decides to pack the new product in closed right circular cylindrical cans of volume 432 πcm³. The cans are to be made from tin sheet. The company tried different options. Based on the above information, answer the following questions: (i) If 7 cm is the radius of the base of the cylinder and h cm is height, then (a) rh=216 (b) r(r+h)=216 (c) rh² = 432 (d) r 2 h=432 (ii) If S cm² is the surface area of the closed cylindrical can, then (a) S= 2π (r 2 +432/r) (b) S= π (r 2 +864/r) (c) S= π (r 2 +432/r) (d) S= 432π/r (iii) For S to be minimum r is equal to (a) 3 cm (b) 6 cm (c) 8 cm (d) 12 cm (iv) Minimum surface area of cylindrical can is (a) 54 πcm 2 (b) 108 πcm² (c) 216 πcm² (d) none of these (v) The relation between the radius and the height of cylindrical area is (a) height is equal to radius of base (b) height is equal to twice the radius of base (c) radius is equal to twice the height (d) radius is two-third of the height.

Case-study-4 A factory owner wants to construct a tank with rectangular base and rectangular sides, open at the top, so that its depth is 2 m and capacity is 8 m³. The building of the tank costs ₹250 per square metre for the base and ₹180 per square metre for the sides. Based on the above information, answer the following questions: (i) If the length and the breadth of the rectangular base of the tank are x metres and mimiy metres respectively, then the relation between x and y is (a)x+y=4 (b) xy=4 (c) xy=8 (d) xy+x+y=4 (ii) The cost of construction of the sides of the tank is (a) ₹180 (x+y) (b) ₹360 (x+y) (c) ₹720 (x + y) (d) ₹1120 (x+y) (iii) If C (in ₹) is the cost of construction of the tank, then C as a function of x is (d) C= 1120 + 720 (x+4/x) (b) C=720+1120 (x+4/x) (c) C=1120+360 (x+4/x) (d) C=1120+180 (x+4/x) (iv) The cost of construction of the tank is least when the value of x is (a) 1 (b) 3 /2 (c) 2 (d) 3 (v) The least cost of construction of the tank is (a) ₹2000 (b) ₹3000 (c)₹3600 (d) ₹4000

Case-study-5 A carpenter has a wire of length 28 m. He wants to cut into two pieces, one of the two pieces is to be made into a square and other into a circle. Based on the above information, answer the following questions: (i)If x metres wire is used in making a square, then what is the expression of combined area A? (ii) What is the length of radius for minimum combined area? (a) 28/π+4 (b) 112/π+4 (c) 14/π+4 (d) none of these (iii) What is the length of circular part? (a) 28/π+4 (b) 14/π+4 (c) 14π/π+4 (d) 28π/π+4 (iv) What is the length of square part? (a) 112/π+4 (b) 112π/π+4 (c) 14π/π+4 (d) 28/π+4 (v) The minimum combined area of square and circle is (a) 196/π+4 m 2 (b) 196/(π+4) 2 m 2 (c) 196/π-4 m 2 (d) none of these

Case-study-6 A firm has the cost function C(x)=x 3 /3-7x 2 + 111x +50 and demand function x = 100-p. Based on the above information, answer the following questions: (i) The total revenue function is (a) R(x) = x²-100x (b) R(x) = 100x-x 2 (c) R(x) = 100-x (d) none of these (ii) The total profit function is (a)-x 3 /3 +6x²-11x-50 (b)-x 3 /3-6x²-11x+50 (c)x 3 /3+6x²-11x+50 (d) -x 3 /3-6x²-11x+50 (iii) The value of x for which profit is maximum is (a) 8 (b) 9 (c) 10 (d) 11 (iv) The maximum profit is (a) ₹133.11 (b) ₹113.31 (c) ₹111.33 (d) ₹133.11 (b) The marginal revenue when x = 10 units is (a) 900 (b) 80 (c) 90 (d) 800

Case-study-7 The average cost function associated with producing and marketing x units of an item is given by 50 AC=2x-11 + 50/x Based on the above information, answer the following questions: (i) The total cost function is (a) C(x)=2x²-11x+50 (b) C(x)=2- 50/x 2 (c) C(x)=2x- 50/x (d) C(x)=2x²+11x-50 (ii) The marginal cost function is (a) MC=4x+ 11 (b) MC=4x-11 (c) MC = 2 + 50/x 2 (d) MC= 100/x 3 (iii) The marginal cost when x = 4 units is (a) ₹5 (b) ₹27 (c) ₹18 (d) ₹13 (iv) The range of values of x for which AC is increasing is (a) x (b) x>-5 (c) x (d) x>5 (v) The range of values of x for which total cost is decreasing is (a) x ≤ 11/4 (b) x ≥ 11/4 (c) x (d) x>-11/4

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Chapter 6 Class 12 Application of Derivatives

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Learn Chapter 6 Application of Derivatives (AOD) of Class 12 free with solutions of all NCERT Questions for Maths Boards

We learned Derivatives in the last chapter, in Chapter 5 Class 12. In this Chapter we will learn the applications of those derivatives.

The topics in the chapter include

• Finding rate of change
• Checking if a function is increasing or decreasing in an interval
• Checking if a function is increasing or decreasing in whole domain
• Finding if function is strictly increasing or decreasing in an interval
• Finding equation (and slope) of tangent and normal using derivatives
• Finding approximate value of numbers and functions
• Finding minimum and maximum values from graph of a function
• Definition of - Maxima, Minima, Absolute Maxima, Absolute Minima, Point of Inflexion
• Finding local maxima, local minima using first derivative test
• Finding local maxima, local minima using second derivative test
• Finding maximum and minimum values in closed interval ( Finding Absolute Maxima and Absolute Minima )
• Statement questions of maxima and minima where we form equations and solve

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