Da Nang-11/2013 Natural Science Department – Duy Tan University Lecturer: Ho Xuan Binh Optimization Problems. In this section, we will learn: How to solve problems involving maximization and minimization of factors.

Natural Science Department – Duy Tan University Optimization businessperson wants to minimize costs and maximize traveler wants to minimize transportation Principle in optics states that light follows the path that takes the least time. Optimization Problems

Optimization problems STEPS IN SOLVING OPTIMIZATION PROBLEMS Natural Science Department – Duy Tan University Understand the problem 1 Draw a diagram 2 Introduce notation 3 Express F (quantity is to be maximized or minimized) in terms of the variables 4 Express F in terms of one variable 5 Find the absolute Max./Min. Value of F 6

EXAMPLE Natural Science Department – Duy Tan University Optimization problems A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?

Natural Science Department – Duy Tan University Optimization problems Understand the problem is the unknown? What are the depth and width of the field that has the largest are the given quantities? A farmer has 2400 ft of are the given conditions? A farmer wants to fence off a rectangular field that borders a straight river. He needs no fence along the river.

Draw a diagram 2 Natural Science Department – Duy Tan University Optimization problems

Natural Science Department – Duy Tan University Introduce notation 3 Optimization problems We wish to maximize the area A of the rectangle. Let x and y be the depth and width of the rectangle (in feet).

Natural Science Department – Duy Tan University Express A (quantity is to be maximized ) in terms of the variables 4 We express A in terms of x and y: A = xy Optimization problems

Natural Science Department – Duy Tan University Optimization problems Express A in terms of one variable 5 To do this, we use the given information that the total length of the fencing is 2400 ft. Thus, 2x + y = 2400 From that equation, we have: y = 2400 – 2x This gives: A = x(2400 – 2x) = 2400x - 2x 2 Note that 0 ≤ x ≤ 1200 (otherwise A < 0).

Natural Science Department – Duy Tan University Find the absolute Max. Value of A 6 The derivative is: A’(x) = 2400 – 4x So, to find the critical numbers, we solve: 2400 – 4x = 0 This gives: x = 600 A(0) = 0; A(600) = 720,000; and A(1200) = 0 So, the Closed Interval Method gives the maximum value as: A(600) = 720,000 Thus, the rectangular field should be: 600 ft deep, 1200 ft wide Optimization problems

LOGO Thank you for your attention