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4.3: Optimization

Difficulty Level: At Grade Created by: CK-12
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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 \begin{align*}f'(x) = 0\end{align*}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 \begin{align*}F6\end{align*}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 \begin{align*}y = 4x + 7\end{align*}y=4x+7. They need to minimize the function \begin{align*}s = \sqrt{x^2 + y^2}\end{align*}s=x2+y2 where \begin{align*}x\end{align*}x and \begin{align*}y\end{align*}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: \begin{align*}s = \sqrt{x^2 + (4x + 7)^2} = \sqrt{17x^2 + 56x + 49}\end{align*}s=x2+(4x+7)2=17x2+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 \begin{align*}x = - \frac{28}{17}\end{align*}x=2817.

To find the \begin{align*}y-\end{align*}ycoordinate, students should substitute the value of \begin{align*}x\end{align*}x into the original equation \begin{align*}y = 4x + 7\end{align*}y=4x+7. To find the distance, they should substitute the \begin{align*}x-\end{align*}x and \begin{align*}y-\end{align*}yvalues into the function \begin{align*}s = \sqrt{x^2 + y^2}\end{align*}s=x2+y2. The point is (-1.647, 0.412) and the distance is 1.698 units.

Students are to maximize the function \begin{align*}A = l \cdot w\end{align*}A=lw. The constraint is \begin{align*}2l + 2w = 200\end{align*}2l+2w=200. Since \begin{align*}l = 100 - w\end{align*}l=100w students can rewrite the function as \begin{align*}A = (100 - w)w = 100w - w^2\end{align*}A=(100w)w=100ww2.

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

The maximum area is \begin{align*}2500 \ m^2\end{align*}2500 m2.

Problem 2 – Optimization of time derivative problems

Remind students to use \begin{align*}t\end{align*}t, for time, instead of \begin{align*}x\end{align*}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 \begin{align*}y = 20t\end{align*}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 \begin{align*}x = 15 - 15t\end{align*}x=1515t.

Students are to minimize the distance function \begin{align*}s = \sqrt{x^2 + y^2} = \sqrt{(15 - 15t)^2 + (20t)^2}\end{align*}s=x2+y2=(1515t)2+(20t)2.

There is a restriction of \begin{align*}0 < t < 1\end{align*}0<t<1 because the boats are only moving for 1 hour. Students will solve the first derivate to find the critical time is \begin{align*}t = \frac{9}{25}\end{align*}t=925. Since the second derivative is always positive, there is a minimum.

The time at which the distance between the boats is minimized is \begin{align*}\left (\frac{9}{25}\right ) \cdot 60 = 21.6\end{align*}(925)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 \begin{align*} \sin(30^\circ) = 0.5\end{align*} and \begin{align*} \cos(30^\circ) = \frac{\sqrt{3}}{2}\end{align*}.

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

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