3.7lecture
- 1. CHAPTER 3 SECTION 3.7 OPTIMIZATION PROBLEMS
- 2. Applying Our Concepts • We know about max and min … • Now how can we use those principles?
- 4. Use the Strategy • What is the quantity to be optimized? – The volume • What are the measurements (in terms of x)? • What is the variable which will manipulated to determine the optimum volume? 60” • Now use calculus x principles 30”
- 5. Guidelines for Solving Applied Minimum and Maximum Problems
- 6. Optimization
- 7. Optimization Maximizing or minimizing a quantity based on a given situation Requires two equations: Primary Equation what is being maximized or minimized Secondary Equation gives a relationship between variables
- 8. To find the maximum (or minimum) value of a function: 1 Write it in terms of one variable. 2 Find the first derivative and set it equal to zero. 3 Check the end points if necessary.
- 9. Ex. 1 A manufacturer wants to design an open box having a square base and a surface area of 108 in2. What dimensions will produce a box with maximum volume? Since the box has a square h base, its volume is V = x 2h x x Note: We call this the primary equation because it gives a formula for the quantity we wish to optimize. The surface area = the area of the base + the area of the 4 sides. S.A. = x2 + 4xh = 108 We want to maximize the volume, so express it as a function of just one variable. To do this, solve x2 + 4xh = 108 for h.
- 10. h V 108 x 2 4x 2 xh Substitute this into the Volume equation. 108 x x 4x 2 2 27 x x3 4 To maximize V we find the derivative and it’s C.N.’s. dV dx 27 3x 2 4 0 3x2 = 108 C.N.' s x We can conclude that V is a maximum when x = 6 and the dimensions of the box are 6 in. x 6 in. x 3 in. 6
- 11. 2. Find the point on f x x 2 that is closest to (0,3).
- 12. x 2 that is closest to (0,3). Find the point on f x 2. Minimize distance Secondary Primary d x 0 d x 2 2 y 3 y 3 y x2 2 2 ***The value of the root will be smallest when what is inside the root is smallest. d d x d x x 2 y 3 2 2 2 Intervals: 2 x x 3 x 2 x 4 6x 2 9 d x x 4 5x 2 9 d ' x 4x3 10x 4x 3 10x 0 2x 2x 2 5 0 x 0 2x 2 5 0 x 0 x 5 2 Test values: 5 2 , 5 2 ,0 0, 5 2 5 2 , 3 1 1 3 dec inc dec inc d ’(test pt) d(x) rel max rel min x 5 2 5 5 2 2 , rel min x 5 2 5 5 2 2 ,
- 13. 2. A rectangular page is to contain 24 square inches of print. The margins at the top and bottom are 1.5 inches. The margins on each side are 1 inch. What should the dimensions of the print be to use the least paper?
- 14. 2. A rectangular page is to contain 24 square inches of print. The margins at the top and bottom are 1.5 inches. The margins on each side are 1 inch. What should the dimensions of the print be to use the least paper? Primary A A( x ) Secondary xy 24 x 2 y 3 x 2 24 x 48 x 3x 48 x 1 A '( x ) 3 y 3 24 3 x 1.5 6 30 48 x2 y y 2 24 x 24 in 1 24 4 y 3 y 1 x 1.5 6 x 2 Smallest Largest (x is near zero) x 0 crit #'s: x 0, 4 (y is near zero) x 24 Page dimensions: 9 in x 6 in 0,4 4,24 Test values: Print dimensions: 6 in x 4 in Intervals: 1 10 dec inc A ’(test pt) A(x) rel min x 4
- 15. 1. Find two positive numbers whose sum is 36 and whose product is a maximum.
- 16. 1. Find two positive numbers whose sum is 36 and whose product is a maximum. Primary P P x Secondary xy x 36 x y P '( x ) 36 2x 36 2x 0 x 18 Intervals: Test values: 0,18 1 18,36 inc dec 20 P ’(test pt) P(x) rel max x 18 x y 36 y 36 x 36 18 18 18,18
- 17. A Classic Problem You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose? A x 40 2x x x A 40 x 2 x 2 A 40 2x w l x 40 2x 0 40 4x w 10 ft l 40 4x 20 ft 4x 40 x 10 There must be a local maximum here, since the endpoints are minimums.
- 18. A Classic Problem You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose? A x 40 2x x x A 40 x 2 x 2 A 40 2x w l x 40 2x A 10 40 2 10 0 40 4x w 10 ft l 40 4x 20 ft 4x 40 x 10 A 10 20 A 200 ft 2
- 19. Example 5: What dimensions for a one liter cylindrical can will use the least amount of material? We can minimize the material by minimizing the area. We need another equation that relates r and h: V r 2h 3 1 L 1000 cm 1000 1000 r2 r 2h A 2 r 2 2 rh area of ends A 2 r lateral area 1000 2 r r2 2 A 2 r 2 h A 4 r 2000 r 2000 r2
- 20. Example 5: What dimensions for a one liter cylindrical can will use the least amount of material? r 2h V A 2 r 2 2 rh 3 1 L 1000 cm 1000 1000 r2 r 2h lateral area 1000 2 r r2 A 2 r2 h A 2 r 1000 5.42 area of ends 2 h h 10.83 cm A 0 2 4 r 4 r 2000 r2 4 r 2000 4 r 3 500 2000 r 2000 r2 2000 r2 r3 500 r 3 r 5.42 cm
- 21. Notes: If the function that you want to optimize has more than one variable, use substitution to rewrite the function. If you are not sure that the extreme you’ve found is a maximum or a minimum, you have to check. If the end points could be the maximum or minimum, you have to check.
- 22. Example #1 • A company needs to construct a cylindrical container that will hold 100cm3. The cost for the top and bottom of the can is 3 times the cost for the sides. What dimensions are necessary to minimize the cost. 2 r V h r h SA 2 rh 2 r 2
- 23. Minimizing Cost V r 2h 100 SA 2 rh 2 r 2 r SA 2 r SA r 2h 100 100 r 2 2 r 200 2 r2 r Domain: r>0 2 C (r ) C (r ) h 2 200 6 r2 r 200 r 2 12 r
- 24. Minimizing Cost C (r ) 0 200 r2 200 r 200 r2 2 3 r C (1.744 ) 0 12 Concave up – Relative min ------ +++++ 12 r 1.744 0 C' changes from neg. to pos. 200 12 r 3 3 200 12 r 12 r 12 r 100 C (r ) r 3 50 3 1.744 100 r 2 Rel. min h h 10.464 The container will have a radius of 1.744 cm and a height of 10.464 cm