4.3: Optimization
This activity is intended to supplement Calculus, Chapter 3, Lesson 7.
ID: 9609
Time required: 45 minutes
Activity Overview
Students will learn how to use the second derivative test to find maxima and minima in word problems and solve optimization problems in parametric functions.
Topic: Application of Derivatives
- Find the maximum or minimum value of a function in an optimization problem by finding its critical points and applying the second derivative test. Use Solve (in the Algebra menu) to check the solution to .
- Use the command fMin or fMax to verify a manually computed extremum.
- Solve optimization problems involving parametric functions.
Teacher Preparation and Notes
- This investigation uses FMax and fMin to answer questions. Students will have to take derivatives and solve on their own.
- Before starting this activity, students should go to the home screen and select :Clean Up > 2:NewProb, then press ENTER. This will clear any stored variables, turn off any functions and plots, and clear the drawing and home screens.
- This activity is designed to be student-centered with the teacher acting as a facilitator while students work cooperatively.
Associated Materials
- Student Worksheet: Optimization http://www.ck12.org/flexr/chapter/9728, scroll down to third activity.
Problem 1 – Optimization of distance and area
Students will graph the equation . They need to minimize the function where and are the coordinates of a point on the line. The constraint is the equation of the line.
They are to rewrite the function using one variable: .
To find the exact coordinates of the point, students will take the first derivative (Menu > Calculus > Derivative), and solve for the critical value (Menu > Algebra > Solve), and take the second derivative. Since the second derivative is always positive, there a minimum at the critical value of .
To find the coordinate, students should substitute the value of into the original equation . To find the distance, they should substitute the and values into the function . The point is (-1.647, 0.412) and the distance is 1.698 units.
Students are to maximize the function . The constraint is . Since students can rewrite the function as .
Students will take the first derivative and solve to find the critical value is . The second derivative is always negative so we have a maximum. When , then .
The maximum area is .
Problem 2 – Optimization of time derivative problems
Remind students to use , for time, instead of . The position equations are the constraints.
The boat heading north is going from the right angle to the point northward. Its position equation is .
The boat heading west is going to the right angle. At 1 pm, it is one hour from the arrival time 2 pm so it is 15 km away. Its position equation is .
Students are to minimize the distance function .
There is a restriction of because the boats are only moving for 1 hour. Students will solve the first derivate to find the critical time is . Since the second derivative is always positive, there is a minimum.
The time at which the distance between the boats is minimized is minutes after 1 pm or about 1:22 pm. The distance between the two boats is 12 km.
Extension – Parametric Function
To rewrite the parametric equations, students will need to know that and .
To find when the projectile hits the ground, students are to set and solve. ( and ). Substituting these values into the function gives how far away it lands (22,092.5 units). Students can find the maximum height when . Substituting this value in the function for students should get 3188.78 units high.
Image Attributions
Description
Authors:
Tags:
Categories:
Date Created:
Feb 23, 2012Last Modified:
Feb 23, 2012If you would like to associate files with this None, please make a copy first.