Quadratic Functions Min Max
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The maximum area for Betty's garden is 156.25 sq meters when each side is 12.5 meters long. This is found by modeling the problem with the equations: 2L + 2W = 50 MAX = LW Solving the equations gives L = W = 12.5 meters.
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2. Various word problems involving quadratic equations are presented, such as finding unknown lengths, areas, and numbers that satisfy given conditions.
3. The examples are solved step-by-step showing the calculations and reasoning involved in arriving at the solutions.
Calculus1.key
This document presents a new method for solving the classic box-folding problem of determining the largest box that can be made by cutting squares from the corners of a sheet of paper and connecting the edges. The traditional method is described along with its history. A new volume equation is then derived that accounts for additional volume gained by folding the cut-off squares. For a sheet that is a square, this new method provides a 10% larger box volume than the classic approach.
Sample question paper 2 with solution
This document provides a sample question paper for Class XII Mathematics. It consists of 3 sections - Section A has 10 one-mark questions, Section B has 12 four-mark questions, and Section C has 7 six-mark questions. The paper is for 3 hours and carries a total of 100 marks. Some questions provide internal choices. Calculators are not permitted. Sample questions include finding inverse functions, evaluating integrals, solving differential equations, and probability questions.
MA8353 TPDE
1. The document provides 14 problems involving partial differential equations (PDEs). The problems involve forming PDEs by eliminating arbitrary constants from functions, finding complete integrals, and solving PDEs.
2. Methods used include taking partial derivatives, finding auxiliary equations, and making substitutions to isolate the PDE or solve it.
3. The document covers a range of techniques for working with PDEs, including eliminating constants, finding trial solutions, integrating subsidiary equations, and solving auxiliary equations to find complete integrals.
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1) The document discusses completing the square, which involves rewriting quadratic expressions in the form (x + a)2 + b to find maximum and minimum values.
2) Examples are provided of completing the square for expressions like x2 + 8x + 3 and 2x2 + 4x + 11.
3) The technique of setting the expression equal to 0 and solving for x is described as a way to find the minimum value and the corresponding x-value that produces it.
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Quadratic Functions Min Max
- 1. Quadratic Functions - Min/Max Maximum Minimum
- 2. What is the maximum value of the quadratic function: 2 y = -2(x + 4) - 9 (-4, -9) What is the minimum value of the quadratic function: 2 y = 3(x - 8) + 14 (8, 14)
- 3. Betty wants to make a rectangle shaped garden. She has 50 meters of fencing. What lengths of sides give Betty a garden with the maximum area? L W 2L + 2W = 50 MAX = LW
- 4. L 2L + 2W = 50 W MAX = LW L = 50 - 2W = 25 - W 2 MAX = (25 - W)W 2 MAX = -W + 25W
- 5. L 2L + 2W = 50 W MAX = LW 2 MAX = -W + 25W 2 MAX = -{W - 25W} 2 2 2 MAX = -{W - 25W + (25/2) - (25/2) } 2 2 MAX = -{(W - 25/2) - (25/2) }
- 6. L 2L + 2W = 50 W MAX = LW 2 2 MAX = -{(W - 25/2) - (25/2) } 2 MAX = -(W - 12.5) + 156.25
- 7. REMINDER: 2 y = a(x - p) + q The value of y is a maximum if a is - The value of y is a maximum when x = - p The value of y is a minimum if a is + The value of y is a minimum when x = -p
- 8. L 2L + 2W = 50 W MAX = LW 2 MAX = -(W - 12.5) + 156.25 - means a maximum does exist The maximum value will be the vertex, or when W = 12.5, it is 156.25.
- 9. L 2L + 2W = 50 W MAX = LW The maximum value will be the vertex, or when W = 12.5, it is 156.25. Can substitute these values back into the original equations to find the value of L 2L + 2(12.5) = 50 L = 50 - 25 = 12.5 2
- 10. We calculated that the maximum area that can be enclosed with 50 meters 2 of fence is 156.25 m . This area is possible when each side is 12.5m long. 12.5m 12.5m 2 12.5m 156.25m 12.5m
- 11. 304 people will go to a basketball game if the tickets cost 8 dollars. Every time the price is increase $0.50 16 fewer people will go to the game. What ticket price gives the maximum profits?
- 12. Profit = (Number of people)*(Cost per person) P=N*C x = number of times the price is increased P =N *C m m m Nm = N - 16x Cm = C + .5x Pm = (N - 16x) (C + .5x)
- 13. Pm = (N - 16x) (C + .5x) Pm = (304 - 16x) (8 + .5x) P = 2432 + 152x - 128x -8x2 m Pm = -8x2 + 24x + 2432
- 14. 2 + 24x + 2432 Pm = -8x 2 - 3x} Pm = -8{x + 2432 2 - 3x + 1.52 - 1.52} +2432 Pm = -8{x 2 2 Pm = -8{(x - 1.5) -1.5 } + 2432 Pm = -8(x-1.5) 2 + 18 + 2432
- 15. P = -8(x-1.5) 2 + 18 + 2432 m P = -8(x-1.5) 2 + 2450 m So, the max profit is when x = 1.5 x = number of times the price is increased Cm = C + .5x Cm = 8 + .5 (1.5) Cm = $8.75
- 16. ex. 6 1-5