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INSTRUMENTAL
UNDERSTANDING
P=Perimeter
=2x + 2y = 30
A=Area
=xy + x(y-x)
=2xy - x
2
Isolate y in Perimeter Formula
30 = 2x + 2y
2y = 30 – 2x
y = 15 – x
Eliminate y by substituting y from
Perimeter equation into Area
equation.
A = - x
2
+ 2xy
= - x
2
+ 2x(15 – x)
= -3 x2
+ 30x
= -3( x2
- 10x)
Complete the Square
A = -3( x2
- 10x + 25) + 75
= -3 (x−5)2
+ 75
Interpret the Equation:
A parabola that opens down with
vertex at (5,75) so the maximum
area is 75 m2
and this occurs when
x = 5.
Use the Perimeter formula to
solve for y
P = 2x + 2y
30 = 2(5) + 2y
20 = 2y
y = 10
Conclude:
The maximum area is 75 m2
and
is obtained when x = 5 and y = 10
Differentiate Area to show that
(5,75) is a maximum:
A(x) = -3 (x−5)2
+ 75
A'(x) = -3(2)(x-5)
A'(x) = -6x + 30
Solve for the x intercept:
A'(x) = -6x + 30
0 = -6x + 30
6x = 30
x = 5
Choose a value to the left of x:
x = 4
A'(x) = -6x + 30
A'(x) = -6(4) + 30
A'(x) = 6 (positive)
The parabola is increasing to the
left of the vertex.
Choose a value to the right of x:
x = 6
A'(x) = -6x + 30
A'(x) = -6(6) + 30
A'(x) = -6 (negative)
The parabola is decreasing to the
right of the vertex.
Conclude:
Because the parabola increases to
the left of the vertex and
decreases to the right of the
vertex, the vertex is a maximum
not a minimum so the maximum
area occurs at the vertex.
RELATIONAL UNDERSTANDING
INTRODUCE MANIPULATIVES
LET STUDENTS ENTER THE
PROBLEM BY LOOKING AT
SPECIFIC CASES
X Area
1 27
2 48
3 63
4 72
5 75
6 72
7 63
8 48
9 27
SET UP THE
GENERALIZATION
LET STUDENTS WORK IN
GROUPS FROM HERE
WHEN SHARING
SOLUTIONS AS A CLASS:
Point out that technology can be
used so that we do not have to
complete the square, interpret
the parabolic equation, solve for
y or differentiate.
Graph
A = -3 (x−5)2
+ 75
FOR STUDENTS THAT
PLOTTED THE
DERIVATIVE:
x f'(x)
1 24
2 18
3 12
4 6
Use Meta-Calculator to find the
derivative and graph it
Have students make
connections between the
original graph and its derivative
Learner Outcomes
Math 20-1 Solve problems that
involve quadratic equations
(Program of Studies, p.18)
Math 31 Deriving f'(x) for
polynomial functions up to the
third degree, using the definition
of the derivative (Program of
Studies, p.19)
Analyze and synthesize
information to determine patterns
and links among ideas (Program
of Studies, ICT Outcome 4.2)
Nature of Mathematics Patterns
may be represented in concrete,
visual, auditory or symbolic form
(Program of Studies, p.7)
Mathematical Processes Number
Visualization & Technology
(Program of Studies, p.7);
Problem Solving (Program of
Studies, p.6); Communication &
Connections (Program of Studies,
p.5)
Final technology presentatio

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Final technology presentatio

  • 1.
  • 2. INSTRUMENTAL UNDERSTANDING P=Perimeter =2x + 2y = 30 A=Area =xy + x(y-x) =2xy - x 2 Isolate y in Perimeter Formula 30 = 2x + 2y 2y = 30 – 2x y = 15 – x Eliminate y by substituting y from Perimeter equation into Area
  • 3. equation. A = - x 2 + 2xy = - x 2 + 2x(15 – x) = -3 x2 + 30x = -3( x2 - 10x) Complete the Square A = -3( x2 - 10x + 25) + 75 = -3 (x−5)2 + 75 Interpret the Equation: A parabola that opens down with vertex at (5,75) so the maximum area is 75 m2 and this occurs when x = 5.
  • 4. Use the Perimeter formula to solve for y P = 2x + 2y 30 = 2(5) + 2y 20 = 2y y = 10 Conclude: The maximum area is 75 m2 and is obtained when x = 5 and y = 10 Differentiate Area to show that (5,75) is a maximum: A(x) = -3 (x−5)2 + 75 A'(x) = -3(2)(x-5) A'(x) = -6x + 30
  • 5. Solve for the x intercept: A'(x) = -6x + 30 0 = -6x + 30 6x = 30 x = 5 Choose a value to the left of x: x = 4 A'(x) = -6x + 30 A'(x) = -6(4) + 30 A'(x) = 6 (positive) The parabola is increasing to the left of the vertex. Choose a value to the right of x: x = 6
  • 6. A'(x) = -6x + 30 A'(x) = -6(6) + 30 A'(x) = -6 (negative) The parabola is decreasing to the right of the vertex. Conclude: Because the parabola increases to the left of the vertex and decreases to the right of the vertex, the vertex is a maximum not a minimum so the maximum area occurs at the vertex.
  • 7.
  • 9. LET STUDENTS ENTER THE PROBLEM BY LOOKING AT SPECIFIC CASES X Area 1 27 2 48 3 63 4 72 5 75 6 72 7 63 8 48 9 27
  • 10. SET UP THE GENERALIZATION LET STUDENTS WORK IN GROUPS FROM HERE
  • 11. WHEN SHARING SOLUTIONS AS A CLASS: Point out that technology can be used so that we do not have to complete the square, interpret the parabolic equation, solve for y or differentiate. Graph A = -3 (x−5)2 + 75
  • 12. FOR STUDENTS THAT PLOTTED THE DERIVATIVE: x f'(x) 1 24 2 18 3 12 4 6
  • 13. Use Meta-Calculator to find the derivative and graph it Have students make connections between the original graph and its derivative
  • 14. Learner Outcomes Math 20-1 Solve problems that involve quadratic equations (Program of Studies, p.18) Math 31 Deriving f'(x) for polynomial functions up to the third degree, using the definition of the derivative (Program of Studies, p.19) Analyze and synthesize information to determine patterns and links among ideas (Program of Studies, ICT Outcome 4.2)
  • 15. Nature of Mathematics Patterns may be represented in concrete, visual, auditory or symbolic form (Program of Studies, p.7) Mathematical Processes Number Visualization & Technology (Program of Studies, p.7); Problem Solving (Program of Studies, p.6); Communication & Connections (Program of Studies, p.5)