Lecture 14 - Part 2 (MA111) PDF

Lecture 14 – Part 2 Parametric Differentiation Math 1 (MA111) Lecture # 14 Differentiation and Applications Dr. Ahmad Moursy 16 Math 1 (MA111) Lecture # 13 Differentiation and Applications 3.13 The maximum and minimum values of a function Introduction: One of the mo...

Lecture 14 – Part 2 Parametric Differentiation Math 1 (MA111) Lecture # 14 Differentiation and Applications Dr. Ahmad Moursy 16 Math 1 (MA111) Lecture # 13 Differentiation and Applications 3.13 The maximum and minimum values of a function Introduction: One of the most important applications of calculus of differentiation is to find the best way to perform certain task. This problem is eventually known as the optimization problem. Solving an optimization problem usually reduces to finding the maximum or the minimum value of a function and determine where this value occurs. Our task in this section concentrates on developing some mathematical tools for 2 solving this problem. The existence of abs. max or min. depends on the function and on its domain 3 4  abs. max. or min. may occur at an interior point or at an end point 5 If the domain of the continuous function is not a closed interval, then the function may neither have maximum nor minimum. 6  Horizontal tangent Is it true that every critical point is a point of local max. or min.? 7  Example 1: Find the absolute maximum and minimum values for the function f(x) = x2 on the interval [-2, 1]. Steps of (1) Find the values of the function at all critical points solution (2) Find the values of the function at the end points (3) Compare to get the abs. max. and the abs. min. Solution: The function is differential on the whole domain so the only critical points are where f’(x) = 2x = 0. So we need to check the function at the critical point x = 0 and the two end points x = -2 and x = 1. At x = 0 : f(0) = 0 At x = -2: f(-2) = 4 The function has an absolute At x = 1 : f(1) = 1 maximum 4 at x = -2 The function has an absolute minimum 0 at x = 0 8 Absolute maximum Local maximum Absolute minimum 9  Example 2: Find the absolute maximum and minimum values for the function f(x) = x2/3 0n [-2,3]. Solution: f ( x )  x2 / 3  2 2 f ( x )  x 1 / 3  3 3 3 x  the only critical point is x  0 , where f ( 0 ) does not exist. f( 0 )  0 ,f ( 2 )  ( 2 )2 / 3  3 4 f ( 3 )  ( 3 )2 / 3  3 9  the absolute minimum is 0 at x  0 and the absolute 3 maximum is 9 at x  3. ( 3 ) f ( x )  cosh( 2  x ) on [ 0 , 3 ] x Exercises (4) f( x) e on [ 1, 1 ] 10  optimization problems Example 1: Find two nonnegative numbers whose sum is 20 and whose product is as large as possible. Solution: Let us denote the first number by x, then the second number is 20 - x. The product P of the two numbers is given by: P(x) = x(20 - x)=20x - x2, 0 ≤ x ≤ 20. dP Domain is a For the critical points =0 ⇒ 20−2 x= 0⇒ x= 10 closed interval dx P( 10 )=10 ( 20−10 )=100 For x = 0, P(0) = 0, and for x = 20, P(20) = 0. Therefore, the point of absolute maximum is x =10 at which P(10) =100 Conclusion: The two numbers are equal and each of them is 10 and the maximum product is 100. 11 Example 3: A rectangle is to be inscribed in a semicircle of radius 2 in. What is the largest area the rectangle can have, and what are its dimensions? Domain is a closed interval 12  Solution: Let us denote the length of the rectangle by 2x, then in order to get the maximum area, the two upper corners of the rectangle must lie on the given circle. Hence, the y coordinate of the shown corner is as shown in the figure and hence, the area of the rectangle will be A( x )=2 x× √ 4−x 2=2 x √ 4− x 2 , 0≤x≤2 For the critical points: 2 dA −x 2 8−4 x dA =2 x. +2 √ 4−x = ⇒ =0 ⇒ 4 x 2=8 ⇒ x 2=2⇒ x=± √2 dx √ 4−x 2 √ 4 −x 2 dx x=− √2 is rejected because it is out of the domain. So, the only interior critical point is x= √ 2. A( √2 )=4. At x=0, A (0 )=0, and at x=2, A( 2)=0. Therefore, the absolute max. is 4 and it occurs at x= √ 2. Conclusion: The maximum area is 4 square units, for the rectangle height √ 4− x 2 =√ 2 and length 2 x=2 √ 2 units. 13 14 15 16 17 18 19 20 21 22  Can we find ABSOLUTE Exrema on open interval or infinite intervals? Dr. Ahmad Moursy 25