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

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1 Two Parabolas Time required 90 minutes Teaching Goals:. Students interpret the given word problem and complete geometric constructions according to the condition of the problem.. Students choose an independent variable and define it as a constraint in the geometric construction. 3. Students analyze epressions for the optimized quantities, such as length, sum of lengths (perimeter), and area by doing the following: o etermine the domain of calculated functions based on conditions of the problem Prior Knowledge o Use the software to determine visually if a quantity reaches a minimum or maimum o Graph an epression as a function of the chosen independent variable to determine the eistence of a minimum or maimum o etermine epressions for the etrema of calculated functions with the help of the software o onfirm solutions for etrema analytically 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 verte of a parabola. Students should be able to solve quadratic equations. Problem: Given two parabolas defined by the equations y = b and y = b. onsider the region enclosed by these two curves. rectangle with sides parallel to the parabolas ais of symmetry is inscribed in this region. a. Find the etremal perimeter of this rectangle. etermine if this is a minimum or a maimum. b. Find the etremal length of the diagonal of this rectangle. etermine if this is a minimum or a maimum c. Will the area of this rectangle be etremal when perimeter and / or diagonal are etremal?

2 Part 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 epression 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.. Open a new file. If aes do not appear in the blank document, create them by clicking the Toggle grid and aes icon on the top toolbar.. Select Function from the raw toolbo. In the Function Type dialog enter X^ b. The parabola will appear on the graph. Repeat this step and enter function b X^. If needed adjust the window using one of the scaling options. 3. Select Point from the raw toolbo and draw two points on each parabola, one in each quadrant. 4. lick a point in the st quadrant on the upper parabola and on the plot of the parabola. With both of them selected, choose Point proportional along curve from the onstrain toolbo and enter for the value. Here is the -coordinate of the point on the parabola. 5. lick a point in the nd quadrant on the upper parabola and on the plot of the parabola. With both of them selected, choose Point proportional along curve from the onstrain toolbo and enter - for the value. 6. Repeat the same steps for the points on the lower parabola, choosing constraint for the point in the 4 th quadrant, and for the point in the 3 rd quadrant. 7. Select Polygon from the raw toolbo and draw a rectangle with vertices being these four points.

3 .0.5 Y=-X +b.0 B Y=X -b -.5 Q. rag one of the vertices of the rectangle and observe the changes in the perimeter of the rectangle. o you think the perimeter has a minimum? maimum?. From the observations students may conclude that when vertices of rectangle get closer to the coordinate aes, the perimeter is getting smaller, so students may make a conjecture about eistence of maimum. The eistence 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 eist when vertices collapse on the aes and it becomes a segment. Q. What is the range of values for the independent variable,, if the rectangle is inscribed in the region encompassed by two parabolas?. Since rectangle does not eist when vertices are on the coordinate aes, the domain is ( b,0) U (0, b). Q3. Find an epression for the perimeter of the rectangle using software. Plot this epression as a function of and investigate if this function has an etremal value on the domain.. Students can graph the function and determine that the function s plot has a maimum on the domain. Here are the steps for finding an epression for the perimeter and constructing its graph.. lick the rectangle and choose Perimeter in the alculate (Symbolic) toolbo. The software

4 will produce an epression for the perimeter in terms of b and.. Right click the epression for the perimeter and choose opy s/ String. 3. Select Function from the raw toolbo. In the Function Type dialog, artesian Type, paste the epression 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 and the -ais. With both of them selected choose Perpendicular from the onstruct toolbo. The vertical line through point will be constructed. Students can now move point and observe on the graph where the perimeter has its maimal value. 6 5 Y=4 X + - X + b + X - b b+ + b- Y=-X +b B Y=X -b Q4. What are the values of at the function s maima?. Students can observe values of in the Variables toolbo while moving point and conjecture that = 0.5 and They can also enter the value of 0.5 for in the Variables toolbo and observe that the vertical line passes through the maimum of the perimeter function. Q5. onfirm your prediction algebraically.. Students should use the epression for the perimeter given by the software. ue to symmetry about y-ais, it is sufficient to consider the case when > 0. Since < b, then b > 0, so the epression for the perimeter becomes: P= 4+ (b ) = 4( b). The

5 epression is quadratic, so the maimum is reached at the verte of the parabola. Using the 4 formula for the coordinates of the verte, we find: = =. The value of the nd maimum is 8 =. Q5. Find the maimum perimeter of the rectangle.. To solve this algebraically students can substitute the found value of into the function: P= 4 b = 4b+. ue to symmetry, at = we obtain the same result. Q6. To confirm that this value is the maimal value of the function, plot the point with coordinates,4 b +. Here are the steps of the construction:. Select Point from the raw toolbo and draw a point anywhere in the blank space. Select the point and choose oordinat from the onstrain toolbo.. Enter / for 0 and 4* b + for y 0. Observe the point jumping to the function maimum. 5 F Y=4 X + - X + b + X - b 4 3,+4 b b+ + b- Y=-X +b B Y=X -b

6 Part Optimizing the iagonal of the Rectangle In this part of the problem students eplore how the diagonal of the rectangle changes using the dynamic features of the software first, then they can determine the symbolic epression for the diagonal of the rectangle and analyze the function graphically and analytically. The teacher may ask students to save their eisting 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 7 from Part. Here are the steps of construction and questions from this point forward. Select Line Segment from the raw toolbo. raw a segment (connecting the verte in the st quadrant with the verte in the 3 rd quadrant). Q. rag one of the vertices of the rectangle and observe the changes in the length of the diagonal of the rectangle. o you think this length has a minimum? maimum?. From the observations students may conclude that when vertices of the rectangle get closer to the coordinate aes, the diagonal length is getting larger, so students may make a conjecture about eistence of minimum. The maimum does not eist since diagonal is largest when the vertices of the rectangle lie on the y-ais, depending on the value of b, but the rectangle does not eist at these points. Q. Find an epression for the length of the diagonal using the software. Plot this epression as a function of and investigate if this function has an etremal value on the domain.. Students can graph the function and determine that the function plot has a minimum on the domain. Here are the steps for finding an epression for the length of the diagonal and constructing a graph.. lick the diagonal and choose istance/length from the alculate (Symbolic) toolbo. The software will produce an epression for the length of the diagonal in terms of b and.. Right click the epression for the perimeter, choose opy s/ String. 3. Select Function from the raw toolbo. In the Function Type dialog, artesian Type, paste the epression 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.

7 Y= 4 X + X - b b+ - B.0 Y=-X +b Y=X -b Q3. What is the value of where the function has minimum?. Students should use the epression for the length of the diagonal given by the software. 4 onsider the epression under the square root. 4 + ( b) = b+ 4b. This is a biquadratic epression, so we can use z = and write the epression in terms of z: 4( z + ( b) z+ b ). The minimal value is achieved at the verte of the parabola, at z b = =, and thus the function has a minimum at = ± b. Q5. Find the minimum length of the diagonal.. To solve this algebraically students can substitute the value of found above into the function: 4 4 ( ) 4 L= b + b b = b + = b Q6. To confirm that this value is the maimal value of the function, plot the point with coordinates b, 4 b. djust the position of point and confirm your results numerically using.

8 specific values for b. Here are the steps of construction:. Select Point from the raw toolbo and draw a point anywhere in the blank space.. Select the point and choose oordinat from the onstrain toolbo. 3. Enter sqrt( b 0.5) for 0 and sqrt(4* b ) for y 0. Observe the point jumping to the function minimum. 4. Select both point and the -ais. hoose Perpendicular from the onstruct toolbo. The vertical line will appear on the graph. 5. Move point so that the perpendicular intersects the minimum point on the function. Observe the numerical values of b and in the Variables toolbo. Use the formulas that you found to verify that point is at the minimum b, -+4 b b+ Y=-X +b.0 F Y= 4 X + X - b B Y=X -b. For b =, using the formula for we get: = b 0.5 = 0.5 = 0.5 = 0.707, which corresponds the value of = (or a value very close to this) shown in the Variables toolbo with high precision.

9 Part 3 omparison of Points of Etrema for Perimeter, iagonal, and rea. In this part of the problem students investigate if points of etrema 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 epressions for all three quantities and compare graphs of these functions on the domain. The teacher may ask students to save the eisting 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 7 from Part. Here are the steps of construction and questions from this point forward.. Select Line Segment from the raw toolbo. raw a segment (connecting the verte in the st quadrant with the verte in the 3 rd quadrant).. heck the Edit / Settings / Math dialog to make sure that the Use ssumptions value in the Output bo is set to False. (Otherwise your area and perimeter graphs will be quite different.) 3. Plot the epression for the perimeter, diagonal, and area of the rectangle with the following steps: a. lick the object (rectangle or diagonal of the rectangle) and choose the appropriate tool from the alculate (Symbolic) toolbo. The software will produce an epression for the corresponding function in terms of b and. b. Right click the epression for the perimeter, choose opy s / String. c. Select Function from the raw toolbo. In the Function Type dialog, artesian Type, paste the epression 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. hoose different colors for each function plot by right clicking on the plot, and using Properties from the contet menu. 5. Hide epressions for the functions by right clicking the epression and choosing Hide from the contet menu. 6. Select both, point and the -ais. hoose Perpendicular from the onstruct toolbo. vertical line will appear on the graph.

10 5 4 3 B - 4 b b+ + b b+ Q3. oes area of the rectangle have a maimum or minimum on the domain?. s it is seen from the plot of the function, the area has a maimum value at two points that are symmetrical. Q4. re points of etremum for the perimeter and the area the same? For the diagonal and the area?. No, these points do not necessarily coincide. Students can drag point and observe the vertical line through all three functions. Q5. Is it possible for the points of etremum to coincide for all three functions? Eplore using dynamic features of the software.. Students can vary the numerical value of b in the Variables toolbo and try to achieve this condition. The steps of construction are the following:. In the Variables toolbo select parameter b.. Enter a minimum and maimum value for b (in the data entry boes at the bottom left and right corners of the dialog). 3. Use the slider to adjust the value of b and observe the graphs of the functions.

11 4. Move point so that the perpendicular intersects the etremum of one of the functions and compare it with the etrema on the other functions for a particular value of b. Q6. Using epressions for points of etrema found in parts and, find a value of b where the rectangle of maimum perimeter is also the rectangle with the minimum diagonal. Will this rectangle s area be at a maimum?. lgebraically, students can set equal the point of maimum for the perimeter function and the point of minimum for the diagonal length function they found in Parts and of the 3 problem: b =, so b = = Now, they can enter this value for b in the Variables 4 toolbo to see if the area function for this value of b has a maimum at the same point. It appears on the graph that the area does have a maimum at this point. Q7. To confirm that this value is the maimal 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? 3. The real calculation of the slope should have a value of 0, which confirms that when b =, 4 is the value for the maimum area of the rectangle. Here are the steps of construction:. Select Point from the raw toolbo and draw a point on the function of the area.. Select the point and the area function and choose Point proportional along curve from the onstrain toolbo. 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 onstruct toolbo. The tangent line will be constructed. 4. lick the tangent line, choose Slope from the alculate (Real tab) toolbo. The value of the slope will appear on the screen. 5. To avoid accidental changes in the values of b and, enter these values in the Variable toolbo and lock them (select the variable and click the lock icon just below the variables list bo on the right).

12 F b -4 3 B b+ + b b+ -