Case Based Questions - Worksheet 1

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Chapter 6 Class 12 - Application of Derivatives - Case Based Question Worksheet 1 by teachoo Chapter: Chapter 6 Class 12 - Application of Derivatives Name: _____________________________ School: _____________________________ Roll Number: _____________________________ Case Study 1: Optimal Inventory Management A company sells 1200 units of a product per year. The cost of placing one order is ₹100. The cost of holding one unit in inventory for a year is ₹ 0.50 . The company wants to find the optimal order size ( x units) to minimize its total annual inventory cost, which is the sum of ordering costs and holding costs. The average inventory level is x/2. Modeling the Cost: The annual ordering cost is (Total Units / Order Size) * Cost per Order. The annual holding cost is (Average Inventory) * Holding Cost per Unit. Write the total annual inventory cost function, C(x). Finding the Rate of Change: Find the derivative of the cost function, C^' (x). Minimizing Cost: Find the critical point by setting C^' (x)=0. This value of x is the Economic Order Quantity (EOQ). Use the second derivative test to confirm this is a minimum. Practical Application: What is the optimal number of units to order each time, and how many orders will the company place per year to minimize its costs? Case Study 2: Strongest Wooden Beam A cylindrical log has a diameter of 20 cm . A rectangular beam of width w and depth d is to be cut from this log. The strength S of the beam is given by the formula S=kwd^2, where k is a constant. We want to find the dimensions of the strongest possible beam. Constraint Equation: The corners of the rectangular beam must lie on the circular cross-section of the log. Use the Pythagorean theorem to write an equation relating w and d to the diameter of the log. Strength as a Function of One Variable: Use the constraint from part 1 to express the strength S as a function of only one variable (e.g., S(w) ). Maximizing Strength: Find the derivative of the strength function and find the critical value for the width w that maximizes the strength. Optimal Dimensions: What are the width and depth of the strongest beam? How does the ratio d/w relate to the shape of the beam?   Case Study 3: Drone and Observer A drone is rising vertically from a point on the ground at a rate of 5" " m/s. An observer is standing on the ground 120 meters away from the launch point. We want to find how fast the distance between the observer and the drone is changing. Geometric Model: Let y be the height of the drone and z be the distance between the observer and the drone. Draw a right-angled triangle representing the situation and write an equation relating y,z, and the fixed distance of the observer from the launch point. Related Rates Equation: Differentiate the equation from part 1 with respect to time t to get an equation that relates the rates dy/dt and dz/dt. Instantaneous Rate of Change: Find the rate at which the distance z is changing at the instant when the drone is 50 meters high. Limiting Behavior: What happens to the rate of change of the distance, dz/dt, as the drone gets very high? Does it approach a specific value? Important links Answer of this worksheet - https://www.teachoo.com/25586/5359/Case-Based-Questions---Worksheet-1/category/Teachoo-Questions---Case-Based/ Full Chapter with Explanation, Activity, Worksheets and more – https://www.teachoo.com/subjects/cbse-maths/class-12th/ch6-12th-application-of-derivatives/ Maths Class 12 - https://www.teachoo.com/subjects/cbse-maths/class-12th/ For more worksheets, ad-free videos and Sample Papers – subscribe to Teachoo Black here - https://www.teachoo.com/black/   Answer Key to Case Based Question Worksheet 1 Case Study 1: Optimal Inventory Management Cost Function: Ordering Cost =(1200/x)⋅100=120000/x. Holding Cost =(x/2)⋅0.5=0.25x. Total Cost C(x)=120000/x+ 0.25x. Derivative: C^' (x)=-120000/x^2 +0.25. Minimizing Cost: Set C^' (x)=0⇒0.25=120000/x^2 ⇒x^2=120000/0.25=480000⇒x=√480000≈692.8. The optimal order size is approx 693 units. C^'' (x)=240000/x^3 , which is positive for x>0, confirming a minimum. Practical Application: The optimal order size is 693 units. The number of orders per year would be 1200/693≈1.73, meaning they would place 1 or 2 large orders per year. Case Study 2: Strongest Wooden Beam Constraint: By Pythagorean theorem on the circle's cross-section, w^2+d^2=(" diameter " )^2=20^2=400. So, d^2=400-w^2. Strength Function: S(w)=kwd^2=kw(400-w^2 )=400kw-kw^3. Maximizing Strength: S^' (w)=400k-3kw^2=0⇒3w^2=400⇒w=√(400/3)≈11.55" " cm. Optimal Dimensions: w≈11.55" " cm.d^2=400-400/3=800/3⇒d=√(800/3)≈16.33" " cm. The ratio d/w= √(800/3)/√(400/3)=√2. The strongest beam is one where the depth is √2 times the width. Case Study 3: Drone and Observer Geometric Model: The situation forms a right-angled triangle. By Pythagorean theorem, 120^2+y^2=z^2. Related Rates Equation: Differentiating with respect to t:0+2y dy/dt=2z dz/dt⇒dz/dt=y/z dy/dt. Instantaneous Rate: When y=50,z=√(120^2+50^2 )=√(14400+2500)=√16900=130" " m. We are given dy/dt=5. So, dz/dt=50/130⋅5=25/13≈1.92" " m/s. Limiting Behavior: As the drone gets very high, y and z become very large, and the ratio y/z approaches 1 (since the fixed 120 m becomes insignificant). Therefore, dz/dt approaches dy/dt, which is 5" " m/s.