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4.7 FORMING FUNCTIONS FROM VERBAL DESCRIPTIONS

4.7 FORMING FUNCTIONS FROM VERBAL DESCRIPTIONS . Objectives: 1) Form a function in one variable from a verbal description 2) Use the graphing calculator to approximate a value for a graphic solution to a verbal problem . Question 1

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4.7 FORMING FUNCTIONS FROM VERBAL DESCRIPTIONS

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  1. 4.7FORMING FUNCTIONS FROM VERBAL DESCRIPTIONS Objectives: 1) Form a function in one variable from a verbal description 2) Use the graphing calculator to approximate a value for a graphic solution to a verbal problem

  2. Question 1 A pile of sand is in the shape of a cone with a diameter that is twice its height. Sketch the diagram. Express the volume V as a function of the height of the pile. Since the volume of a cone is given by V = 1/3r2h and d = 2h, therefore 2r = 2h or h = r, we have V = 1/3h3 In this last equation, the volume is a function of the height of the sand pile (cone).

  3. Strategy 1) Identify given information. 2) Assign variables to the related quantities. 3) Use formula(s) to link all variables together (Primary Equation). 4) Find out the relationship among other variables to one variable (Secondary Equation(s)). 5) Rewrite the formula(s) again in terms of one variable.

  4. Question 2 A winch at the tip of a 12-meter building pulls pipe of the same length to a vertical position. The length of a rope at top of the building holds the pipe. Find the length of the rope in terms of the height. s (x, y) 12 12 [Solution] The relationships among the three variables are: and

  5. Question 3 An open top box with a square base is to be constructed from sheet metal in such a way that the completed box is made from 2 square yards of sheet metal. • Express the volume of the box as a multivariable function of the base and width of the square base dimensions. • Express the volume of the box as a single function of the width of the square base. • Find the feasible domain of the side of the base. [Solution](1) Let the length of the base be x inch, h and the height be h inch. Then the volume of the box in terms of x andh inch is x x Primary Equation

  6. (2) Notice that “made from a 2 yd2 sheet metal” provides Secondary Equation From the secondary equation, we have Then the volume of the box in terms of x in the primary equation is (3) Since the height h > 0, then we need to solve the rational inequality This results in

  7. Question 4 A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimension will produce a box with maximum volume? Solution Let the length of the base be x inch, h and the height be h inch. Then the volume of the box in terms of x andh inch is x x Primary Equation Notice that “surface area of 108 in2 ” provides Secondary Equation

  8. Question 4 A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimension will produce a box with maximum volume? Solution From the secondary equation, we have h Then the volume of the box in terms of x in the primary equation is x x Now, before we aim on V(x) and try to find its maximum volume while x has certain length, we should find the feasible domain of x.

  9. Notice that and This means Feasible Domain Therefore, V(6, 3) is maximum when x = 6 and h = 3. The box dimension is 6 x 6 x 3.

  10. Question 5 Which points on the graph of y = 4 – x2 are closest to the point (0, 2)? Solution The distance between the point (0, 2) to any point (x, y) on the graph is (x, y) d Notice that y = 4 – x2, then Also notice that the x value makes the d(x) reaches the minimum is the same as makes the d2(x) reaches the minimum.

  11. Question 5 Which points on the graph of y = 4 – x2 are closest to the point (0, 2)? Solution Therefore, When and or The d2(x) reaches the minimum. The minimum value is So the points are closest to (0, 2) and the distanceis

  12. Question 6 From a raft 50 m offshore, a lifeguard wants to swim to shore and run to a snack bar 100 m down the beach, as show. • If the lifeguard swims at 1 m/s and run at 3 m/s, express the total swimming and running time t as a function of the distance x show in the diagram? • Find the minimum time. Solution The swimming distance in terms of x is 50 The running distance in terms of x is x 100 The total time in second in terms of x is

  13. Question 7 A man is in boat 2 miles from the nearest point R on the coast. His is to go to a point Q, located 3 miles down the coast and 1 mile in land. He can row at 2 miles per hour and walk at 4 miles per hour. Toward what point on the coast should he row in order to reach point Q in the least time? Solution M Let the point on the coast toward by the man be P, and PR = x, then PS = 3 – x. Then 2 3 – x P R 3 S x 1 Q

  14. Then the total time from point M to Q will be M 2 3 – x P R 3 S x 1 Q The feasible domain for x is [0, 3]. The solution for x within the domain is x = 1.

  15. Question 8 Four feet wire is to be used to form a square and a circle. How much of the wire should be used for the square and how much should be used for the circle to enclose the maximum area? x Solution x Let the side of the square be x and the radius of the circle be r. Then Primary Equation Since the total length of wire is 4 feet, then r Secondary Equation So,

  16. x x r The feasible domain of x is restricted by the square’s perimeter We conclude that when yields the maximum area. That means all the wire is used to form circle.

  17. B A (x, y) C D • Question 9 The Rectangle ABCD has vertices C and D on the x-axis and vertices A and B on the part of the parabola • y = 4 – x2 that is above the x-axis. • Express the perimeter P of the rectangle as a function of the x-coordinate of A. • What is the domain of the perimeter • function? • c) For what value of x is the perimeter • a maximum? Solution (a)

  18. B A (x, y) C D • Question 9 The Rectangle ABCD has vertices C and D on the x-axis and vertices A and B on the part of the parabola • y = 4 – x2 that is above the x-axis. • Express the perimeter P of the rectangle as a function of the x-coordinate of A. • What is the domain of the perimeter • function? • c) For what value of x is the perimeter • a maximum? Solution (b) Domain of the perimeter is

  19. B A (x, y) C D • Question 9 The Rectangle ABCD has vertices C and D on the x-axis and vertices A and B on the part of the parabola • y = 4 – x2 that is above the x-axis. • Express the perimeter P of the rectangle as a function of the x-coordinate of A. • What is the domain of the perimeter • function? • c) For what value of x is the perimeter • a maximum? Solution (c) From the expression in (a), when x = 1, P(x) has a maximum.

  20. B A (x, y) C D • Question 10 The Rectangle ABCD has vertices C and D on the x-axis and vertices A and B on the part of the parabola • y = 4 – x2 that is above the x-axis. • Express the area A of the rectangle as a function of the x-coordinate of A. • What is the domain of the area • function? • c) For what value of x is the area • a maximum? Solution (a)

  21. B A (x, y) C D • Question 10 The Rectangle ABCD has vertices C and D on the x-axis and vertices A and B on the part of the parabola • y = 4 – x2 that is above the x-axis. • Express the area A of the rectangle as a function of the x-coordinate of A. • What is the domain of the area • function? • c) For what value of x is the area • a maximum? Solution (b) Domain of the area is

  22. B A (x, y) C D • Question 10 The Rectangle ABCD has vertices C and D on the x-axis and vertices A and B on the part of the parabola • y = 4 – x2 that is above the x-axis. • Express the area A of the rectangle as a function of the x-coordinate of A. • What is the domain of the area • function? • c) For what value of x is the area • a maximum? Solution (c) From the expression in (a), when x 1.155, P(x) has a maximum.

  23. AssignmentDay 1P. 150 #16 – 24 (even),P. 161 #2, 3, 5, 7, 8, 10, 11Day 2P. 162 #14, 15, 18, 19, 21, WS 4.3B

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