o Graph an expression as a function of the chosen independent variable to determine the existence of a minimum or maximum

Two Parabolas

_{Time required} 90 minutes Teaching Goals:

1.

Students interpret the given word problem and complete geometric constructions according to the condition of the problem.

2.

Students choose an independent variable and define it as a constraint in the geometric construction.

3.

Students analyze expressions for the optimized quantities, such as length, sum of lengths (perimeter), and area by doing the following:

o

Determine the domain of calculated functions based on conditions of the problem

o

Use the software to determine visually if a quantity reaches a minimum or

maximum

o

Graph an expression as a function of the chosen independent variable to determine the existence of a minimum or maximum

o

Determine expressions for the extrema of calculated functions with the help of the software

o

Confirm solutions for extrema analytically Prior Knowledge

• Students should know the properties of a rectangle.

• Students should know the equation and graph of a parabola.

• Students should know the formula for finding the vertex of a parabola. • Students should be able to solve quadratic equations.

Problem:

Given two parabolas defined by the equations

y

=

x

2

−

b

and

y

= −

b

x

2. Consider the region enclosed by these two curves. A rectangle with sides parallel to the parabolas’ axis of

symmetry is inscribed in this region.

a. Find the extremal perimeter of this rectangle. Determine if this is a minimum or a maximum.

b. Find the extremal length of the diagonal of this rectangle. Determine if this is a minimum or a maximum

c. Will the area of this rectangle be extremal when perimeter and / or diagonal are extremal?

Part 1 – Optimizing Perimeter of Rectangle

In this part of the problem students analyze the word problem and complete the geometric construction of two parabolas and an inscribed rectangle according to the condition of the problem. Students then choose an independent variable and define it in the construction. Students can use the dynamic features of the software to analyze the problem qualitatively first, then they can determine the symbolic expression for the perimeter of the rectangle using the symbolic calculation feature and analyze the problem graphically and analytically. Here are the steps of construction and questions.

1. Open a new file. If axes do not appear in the blank document, create them by clicking the Toggle grid and axes icon on the top toolbar.

2. Select Function from the Draw toolbox. In the Function Type dialog enter X^2 – b. The parabola will appear on the graph. Repeat this step and enter function b – X^2. If needed adjust the window using one of the scaling options.

3. Select Point from the Draw toolbox and draw two points on each parabola, one in each quadrant.

4. Click a point in the 1st quadrant on the upper parabola and on the plot of the parabola. With both of them selected, choose Point proportional along curve from the Constrain toolbox and enter x for the value. Here

x

is the x-coordinate of the point on the parabola.

5. Click a point in the 2nd quadrant on the upper parabola and on the plot of the parabola. With both of them selected, choose Point proportional along curve from the Constrain toolbox and enter -x for the value.

6. Repeat the same steps for the points on the lower parabola, choosing constraint x for the point in the 4th quadrant, and –x for the point in the 3rd quadrant.

7. Select Polygon from the Draw toolbox and draw a rectangle with vertices being these four points.

0.5 1.0 1.5 2.0 2.5 -0.5 -1.0 -1.5 -2.0 -2.5 0.5 1.0 1.5 2.0 -0.5 -1.0 -1.5 C A B D Y=-X2+b x -x x Y=X2-b -x

Q1.

Drag one of the vertices of the rectangle and observe the changes in the perimeter of the rectangle. Do you think the perimeter has a minimum? A maximum?

A

. From the observations students may conclude that when vertices of rectangle get closer to the coordinate axes, the perimeter is getting smaller, so students may make a conjecture about existence of maximum. The existence of minimum cannot be established from the visual

observations due to the fact that it is not clear if the function for the perimeter is monotonic and since the rectangle does not exist when vertices collapse on the axes and it becomes a segment.

Q2.

What is the range of values for the independent variable, x, if the rectangle is inscribed in the region encompassed by two parabolas?

A.

Since rectangle does not exist when vertices are on the coordinate axes, the domain is

(

−

b

, 0)

_U

(0,

b

)

.

Q3.

Find an expression for the perimeter of the rectangle using software. Plot this expression as a function of

x

and investigate if this function has an extremal value on the domain.

A.

Students can graph the function and determine that the function’s plot has a maximum on the domain. Here are the steps for finding an expression for the perimeter and constructing its graph. 1. Click the rectangle and choose Perimeter in the Calculate (Symbolic) toolbox. The software

will produce an expression for the perimeter in terms of

b

and

x

.

2. Right click the expression for the perimeter and choose Copy As/ String.

3. Select Function from the Draw toolbox. In the Function Type dialog, Cartesian Type, paste the expression into the Y= line and press OK. The graph of the function will appear on the screen. If you can’t see the function, choose one of the scaling options to adjust the plot. 4. Select point A and the x-axis. With both of them selected choose Perpendicular from the

Construct toolbox. The vertical line through point A will be constructed. Students can now move point A and observe on the graph where the perimeter has its maximal value.

1 2 3 4 5 -1 -2 -3 -4 -5 1 2 3 4 5 6 -1 C A B D ⇒ 4· x + -2·b+2·x2 + 2·b-2·x2 Y=4· X + -2·X2+2·b + 2·X2-2·b -x x Y=X2-b x -x Y=-X2+b

Q4.

What are the values of

x

at the function’s maxima?

A.

Students can observe values of x in the Variables toolbox while moving point A and conjecture that

x

= 0.5 and -0.5. They can also enter the value of 0.5 for x in the Variables toolbox and observe that the vertical line passes through the maximum of the perimeter function.

Q5.

Confirm your prediction algebraically.

A.

Students should use the expression for the perimeter given by the software. Due to symmetry about

y

-axis, it is sufficient to consider the case when

x

> 0. Since

x

<

b

, then

2

b

−

2

x

2

>

0

, so the expression for the perimeter becomes:

P

=

4

x

+

2(2

b

−

2

x

2

)

= −

4(

x

2

− −

x b

)

. The

expression is quadratic, so the maximum is reached at the vertex of the parabola. Using the formula for the coordinates of the vertex, we find:

4

1

8

2

x

=

−

=

−

. The value of the 2nd maximum is

1

2

x

= −

.

Q5.

Find the maximum perimeter of the rectangle.

A.

To solve this algebraically students can substitute the found value of

x

into the function: 2

1

1

4

4

1

2

2

P

= −

⎛

⎜

_⎜

⎛ ⎞

_{⎜ ⎟}

− −

b

⎞

⎟

_⎟

=

b

+

⎝ ⎠

⎝

⎠

. Due to symmetry, at

1

2

x

= −

we obtain the same result.

Q6.

To confirm that this value is the maximal value of the function, plot the point with coordinates

1

, 4

1

2

b

⎛

₊

⎞

⎜

⎟

⎝

⎠

. Here are the steps of the construction:

1. Select Point from the Draw toolbox and draw a point anywhere in the blank space. Select the point and choose Coordinat from the Constrain toolbox.

2. Enter 1/ 2 for

x

0 and 4 *b+1 for

y

0. Observe the point jumping to the function maximum.

1 2 3 4 5 6 -1 -2 -3 -4 1 2 3 4 5 -1 -2 C A B F D ⇒ 4· x + -2·b+2·x2 + 2·b-2·x2 Y=4· X + -2·X2+2·b + 2·X2-2·b -x x Y=X2-b 1 2,1+4·b x -x Y=-X2+b

Part 2 – Optimizing the Diagonal of the Rectangle

In this part of the problem students explore how the diagonal of the rectangle changes using the dynamic features of the software first, then they can determine the symbolic expression for the diagonal of the rectangle and analyze the function graphically and analytically. The teacher may ask students to save their existing file, including constructions and computations for the

rectangle, and delete all unnecessary constructions and calculations. If the teacher decides to start with the blank file, repeat steps 1 – 7 from Part 1. Here are the steps of construction and questions from this point forward.

Select Line Segment from the Draw toolbox. Draw a segment AC (connecting the vertex in the 1st quadrant with the vertex in the 3rd quadrant).

Q1.

Drag one of the vertices of the rectangle and observe the changes in the length of the diagonal of the rectangle. Do you think this length has a minimum? A maximum?

A

. From the observations students may conclude that when vertices of the rectangle get closer to the coordinate axes, the diagonal length is getting larger, so students may make a conjecture about existence of minimum. The maximum does not exist since diagonal is largest when the vertices of the rectangle lie on the y-axis, depending on the value of

b

, but the rectangle does not exist at these points.

Q2.

Find an expression for the length of the diagonal using the software. Plot this expression as a function of

x

and investigate if this function has an extremal value on the domain.

A.

Students can graph the function and determine that the function plot has a minimum on the domain. Here are the steps for finding an expression for the length of the diagonal and

constructing a graph.

1. Click the diagonal and choose Distance/Length from the Calculate (Symbolic) toolbox. The software will produce an expression for the length of the diagonal in terms of

b

and

x

.

2. Right click the expression for the perimeter, choose Copy As/ String.

3. Select Function from the Draw toolbox. In the Function Type dialog, Cartesian Type, paste the expression into the Y= line and press OK. The graph of the function will appear on the screen. If you can’t see the function, choose one of the scaling options to adjust the plot.

0.5 1.0 1.5 2.0 2.5 -0.5 -1.0 -1.5 -2.0 -2.5 0.5 1.0 1.5 2.0 2.5 -0.5 -1.0 D B C A ⇒ 4·x2 + -2·b+2·x2 2 -x Y= 4·X2+ 2·X2-2·b 2 Y=-X2+b x -x x Y=X2-b

Q3.

What is the value of

x

where the function has minimum?

A.

Students should use the expression for the length of the diagonal given by the software. Consider the expression under the square root.

4

x

2

+

(2

x

2

−

2 )

b

2

=

4

x

2

+

4

x

4

−

8

x b

2

+

4

b

2. This is a biquadratic expression, so we can use

z

=

x

2and write the expression in terms of

z

:

2 2

4(

z

+ −

(1 2 )

b z

+

b

)

. The minimal value is achieved at the vertex of the parabola, at 2

2

1

2

b

z

=

−

=

x

, and thus the function has a minimum at 1 2

x= ± b− .

Q5.

Find the minimum length of the diagonal.

A.

To solve this algebraically students can substitute the value of

x

found above into the function: 2 2

1

1

4

2

2

4

2 ( 1)

4

1

2

2

L

=

⎛

_⎜

b

−

_⎟

⎞

+

⎛

_⎜

⎛

_⎜

b

−

⎞

_⎟

−

b

⎞

_⎟

=

b

− + −

=

b

−

⎝

⎠

⎝

⎝

⎠

⎠

.

Q6.

To confirm that this value is the maximal value of the function, plot the point with coordinates

1

, 4

1

2

b

b

⎛

⎞

−

−

⎜

⎟

⎜

⎟

specific values for

b.

Here are the steps of construction:

1. Select Point from the Draw toolbox and draw a point anywhere in the blank space. 2. Select the point and choose Coordinat from the Constrain toolbox.

3. Enter

sqrt b

(

−

0.5)

for

x

0 and

sqrt

(4 *

b

−

1)

for

y

0. Observe the point jumping to the function minimum.

4. Select both point A and the x-axis. Choose Perpendicular from the Construct toolbox. The vertical line will appear on the graph.

5. Move point A so that the perpendicular intersects the minimum point on the function. Observe the numerical values of

b

and

x

in the Variables toolbox. Use the formulas that you found to verify that point A is at the minimum.

0.5 1.0 1.5 2.0 2.5 -0.5 -1.0 -1.5 -2.0 -2.5 0.5 1.0 1.5 2.0 2.5 -0.5 -1.0 F D B C A

⇒ 4·x

2

+ -2·b+2·x

2 2 -x

-0.5+b, -1+4·b

Y= 4·X

2

+ 2·X

2

-2·b

2 Y=-X2+b x -x x Y=X2-b

A.

For

b

= 1, using the formula for

x

we get:

x

=

b

−

0.5

=

1 0.5

−

=

0.5

=

0.707

, which corresponds the value of

x

= 0.703 (or a value very close to this) shown in the Variables toolbox with high precision.

Part 3 – Comparison of Points of Extrema for Perimeter, Diagonal, and Area.

In this part of the problem students investigate if points of extrema for the perimeter, diagonal, and area of the previously inscribed rectangle can be the same. Using dynamic and symbolic features of the software, students will plot expressions for all three quantities and compare graphs of these functions on the domain. The teacher may ask students to save the existing file that already includes constructions for the rectangle, and delete all unnecessary constructions and calculations. If the teacher decides to start with the blank file, repeat steps 1 – 7 from Part 1. Here are the steps of construction and questions from this point forward.

1. Select Line Segment from the Draw toolbox. Draw a segment AC (connecting the vertex in the 1st quadrant with the vertex in the 3rd quadrant).

2. Check the Edit / Settings / Math dialog to make sure that the Use Assumptions value in the Output box is set to False. (Otherwise your area and perimeter graphs will be quite different.) 3. Plot the expression for the perimeter, diagonal, and area of the rectangle with the following

steps:

a. Click the object (rectangle or diagonal of the rectangle) and choose the appropriate tool from the Calculate (Symbolic) toolbox. The software will produce an expression for the corresponding function in terms of

b

and

x

.

b. Right click the expression for the perimeter, choose Copy As / String.

c. Select Function from the Draw toolbox. In the Function Type dialog, Cartesian Type, paste the expression into the Y= line and press OK. The graph of the function will appear on the screen. If you can’t see the function, choose one of the scaling options to adjust the plot.

4. Choose different colors for each function plot by right clicking on the plot, and using Properties from the context menu.

5. Hide expressions for the functions by right clicking the expression and choosing Hide from the context menu.

6. Select both, point A and the x-axis. Choose Perpendicular from the Construct toolbox. A vertical line will appear on the graph.

1 2 3 4 -1 -2 -3 -4 1 2 3 4 5 -1 D B C A ⇒ 4·x2 + -2·b+2·x2 2 ⇒ 4· x + -2·b+2·x2 + 2·b-2·x2 ⇒ 4·b·x-4·x3 -x x -x x

Q3.

Does area of the rectangle have a maximum or minimum on the domain?

A.

As it is seen from the plot of the function, the area has a maximum value at two points that are symmetrical.

Q4.

Are points of extremum for the perimeter and the area the same? For the diagonal and the area?

A.

No, these points do not necessarily coincide. Students can drag point A and observe the vertical line through all three functions.

Q5.

Is it possible for the points of extremum to coincide for all three functions? Explore using dynamic features of the software.

A1.

Students can vary the numerical value of

b

in the Variables toolbox and try to achieve this condition. The steps of construction are the following:

1. In the Variables toolbox select parameter

b

.

2. Enter a minimum and maximum value for

b

(in the data entry boxes at the bottom left and right corners of the dialog).

4. Move point A so that the perpendicular intersects the extremum of one of the functions and compare it with the extrema on the other functions for a particular value of

b

.

Q6.

Using expressions for points of extrema found in parts 1 and 2, find a value of

b

where the rectangle of maximum perimeter is also the rectangle with the minimum diagonal. Will this rectangle’s area be at a maximum?

A.

Algebraically, students can set equal the point of maximum for the perimeter function and the point of minimum for the diagonal length function they found in Parts 1 and 2 of the

problem: 1 1 2 2

b− = , so

3

0.75

4

b

= =

. Now, they can enter this value for

b

in the Variables toolbox to see if the area function for this value of

b

has a maximum at the same point. It appears on the graph that the area does have a maximum at this point.

Q7.

To confirm that this value is the maximal value of the function, plot a tangent line at this point and determine the slope of the tangent line. What is the value of the slope?

A.

The real calculation of the slope should have a value of 0, which confirms that when

3

4

b

=

,

2

1

is the x value for the maximum area of the rectangle. Here are the steps of construction:

1. Select Point from the Draw toolbox and draw a point on the function of the area.

2. Select the point and the area function and choose Point proportional along curve from the Constrain toolbox. Enter 0.5 for the constraint value and the point should move to the intersection between the graph of the function and the vertical line.

3. Select this point and the plot of the function and choose Tangent from the Construct toolbox. The tangent line will be constructed.

4. Click the tangent line, choose Slope from the Calculate (Real tab) toolbox. The value of the slope will appear on the screen.

5. To avoid accidental changes in the values of

b

and

x

, enter these values in the Variable toolbox and lock them (select the variable and click the lock icon just below the variables list box on the right).

1 2 3 4 -1 -2 -3 -4 1 2 3 4 5 -1 D B F C A

⇒ 4·x

2

+ -2·b+2·x

2 2

⇒ 4· x + -2·b+2·x

2

+ 2·b-2·x

2

⇒ 4·b·x-4·x

3 ⇒ ∼0 x -x x 0.5 -x