Answered: Effect of Price on Supply of Eggs…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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$### Effect of Price on Supply of Eggs Suppose the wholesale price of a certain brand of medium-sized eggs $ p $ (in dollars/carton) is related to the weekly supply $ x $ (in thousands of cartons) by the following equation: \[ 625p^2 - x^2 = 100 \] **Problem Statement:** If 20,000 cartons of eggs are available at the beginning of a certain week and the price is falling at the rate of 6¢/carton/week, at what rate is the supply changing? (Round your answer to the nearest whole number.) *Hint: To find the value of $ p $ when $ x = 20 $, solve the supply equation for $ p $. Input your answer in the provided box in the format: \[ \_\_\_\_ \text{ cartons per week} \] #### Steps to Solve: 1. **First Step: Solving for $ p $** When $ x = 20,000 $, convert $ x $ to thousands of cartons: \[ x = \frac{20,000}{1,000} = 20 \] Substitute $ x = 20 $ into the equation: \[ 625p^2 - 20^2 = 100 \] Simplify and solve for $ p $: \[ 625p^2 - 400 = 100 \] \[ 625p^2 = 500 \] \[ p^2 = \frac{500}{625} \] \[ p^2 = 0.8 \] \[ p = \sqrt{0.8} \] \[ p \approx 0.894 \text{ dollars/carton} \] 2. **Second Step: Using the Rate of Price Change** Given that the price is falling at the rate of 6¢/carton/week: \[ dp/dt = -0.06 \text{ dollars/week} \] Differentiate the supply equation implicitly with respect to time $ t $: \[ 2(625p)\frac{dp}{dt} - 2x\frac{dx}{dt} = 0 \] Simplify: \[ 1250p\frac{dp}{dt} - 2x\frac{dx}{dt$

Transcribed Image Text:### Effect of Price on Supply of Eggs Suppose the wholesale price of a certain brand of medium-sized eggs $ p $ (in dollars/carton) is related to the weekly supply $ x $ (in thousands of cartons) by the following equation: \[ 625p^2 - x^2 = 100 \] **Problem Statement:** If 20,000 cartons of eggs are available at the beginning of a certain week and the price is falling at the rate of 6¢/carton/week, at what rate is the supply changing? (Round your answer to the nearest whole number.) *Hint: To find the value of $ p $ when $ x = 20 $, solve the supply equation for $ p $. Input your answer in the provided box in the format: \[ \_\_\_\_ \text{ cartons per week} \] #### Steps to Solve: 1. **First Step: Solving for $ p $** When $ x = 20,000 $, convert $ x $ to thousands of cartons: \[ x = \frac{20,000}{1,000} = 20 \] Substitute $ x = 20 $ into the equation: \[ 625p^2 - 20^2 = 100 \] Simplify and solve for $ p $: \[ 625p^2 - 400 = 100 \] \[ 625p^2 = 500 \] \[ p^2 = \frac{500}{625} \] \[ p^2 = 0.8 \] \[ p = \sqrt{0.8} \] \[ p \approx 0.894 \text{ dollars/carton} \] 2. **Second Step: Using the Rate of Price Change** Given that the price is falling at the rate of 6¢/carton/week: \[ dp/dt = -0.06 \text{ dollars/week} \] Differentiate the supply equation implicitly with respect to time $ t $: \[ 2(625p)\frac{dp}{dt} - 2x\frac{dx}{dt} = 0 \] Simplify: \[ 1250p\frac{dp}{dt} - 2x\frac{dx}{dt

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