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# 4.3: Optimization

Difficulty Level: At Grade Created by: CK-12

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 $f'(x) = 0$.
• 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 $F6$ :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

## Problem 1 – Optimization of distance and area

Students will graph the equation $y = 4x + 7$. They need to minimize the function $s = \sqrt{x^2 + y^2}$ where $x$ and $y$ 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: $s = \sqrt{x^2 + (4x + 7)^2} = \sqrt{17x^2 + 56x + 49}$.

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 $x = - \frac{28}{17}$.

To find the $y-$coordinate, students should substitute the value of $x$ into the original equation $y = 4x + 7$. To find the distance, they should substitute the $x-$ and $y-$values into the function $s = \sqrt{x^2 + y^2}$. The point is (-1.647, 0.412) and the distance is 1.698 units.

Students are to maximize the function $A = l \cdot w$. The constraint is $2l + 2w = 200$. Since $l = 100 - w$ students can rewrite the function as $A = (100 - w)w = 100w - w^2$.

Students will take the first derivative and solve to find the critical value is $w = 50$. The second derivative is always negative so we have a maximum. When $w = 50 \ m$, then $l = 50 \ m$.

The maximum area is $2500 \ m^2$.

## Problem 2 – Optimization of time derivative problems

Remind students to use $t$, for time, instead of $x$. The position equations are the constraints.

The boat heading north is going from the right angle to the point northward. Its position equation is $y = 20t$.

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 $x = 15 - 15t$.

Students are to minimize the distance function $s = \sqrt{x^2 + y^2} = \sqrt{(15 - 15t)^2 + (20t)^2}$.

There is a restriction of $0 < t < 1$ because the boats are only moving for 1 hour. Students will solve the first derivate to find the critical time is $t = \frac{9}{25}$. Since the second derivative is always positive, there is a minimum.

The time at which the distance between the boats is minimized is $\left (\frac{9}{25}\right ) \cdot 60 = 21.6$ 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 $\sin(30^\circ) = 0.5$ and $\cos(30^\circ) = \frac{\sqrt{3}}{2}$.

To find when the projectile hits the ground, students are to set $y = 0$ and solve. ($t = 0$ and $t = 51.02$). Substituting these values into the $x$ function gives how far away it lands (22,092.5 units). Students can find the maximum height when $\frac{dy}{dt} = 0 \ (t \approx 25.51)$. Substituting this value in the function for $y$ students should get 3188.78 units high.

Feb 23, 2012

Nov 04, 2014