Modeling and Optimization. One of the major roles of Differential Calculus is the determining of the Maximum and Minimum values of a situation.

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1 Modeling and Optimization One of the major roles of Differential Calculus is the determining of the Maimum and Minimum values of a situation. These minimum or maimum (Etreme Values) will occur at the following points if the domain is a b.. The end points f(a) or f(b). Where f 0 Steps in solving Applied Maimum and Minimum Problems. Understand the problem take time to read the question slowly and carefully until it is clearly understood.. Draw a diagram and label (introduce) the unknown variables, given quantities, and given conditions. 3. Introduce Notation for quantity that is to be maimized or minimized. For eample: A for area, T for time, or V for volume.. Note the given quantities and relationships 5. Formulate the relationships between the given variables and the variable that is to be maimized (or minimized). For eample: A r You want to formulate the equation in terms of one variable by using other relationships given or implied by the question. 6. Calculate the first derivative to find the absolute maimum or minimum value of the function by setting the derivative equal to zero (these are the critical points). Don t forget the endpoints (these are valid candidates) 7. Check the critical points and endpoints to determine the maimum or minimum values. Note: Maimum is where f() is the largest Minimum is where f() is the smallest

2 Find the absolute maimum and minimum of the following function Solution f First find the derivative, factor it, set it equal to zero, and solve for f Setting to Zero nd Solving for when =- and =-3 Don t forget the endpoints. The endpoints are =-3.5, and =. We now have all the candidates for the possible maimum and minimum. We only need to check them all in the original function to determine the largest and smallest. Check. f f f f min 8 ma Determine which is the largest and smallest. The function has the maimum value of 8 which occurs when =. The function has the minimum value of which occurs when =-.

3 The concentration C(t) (in milligrams per cubic centimetre) of a certain medicine in a patient's bloodstream is given by C 0.t t 3 t where t is the number of hours after the medicine is taken. Find the maimum and minimum concentrations between the first and sith hours after the medicine has been taken. and solve for t. Set C t 0 0. t 3 0.t t 3 t t 3 0. t 3 0.t t 3 t 3 0. t 3 t 3 t t 3 0. t 3 3 t t t t 3 3 C Therefore C t 0 when t=3. Endpoints The other possibilities are the end points, t= and t=6 Check C C C Therefore the Minimum value of occurs when t= The maimum Value of occurs when t=3

4 The sum of two positive numbers is k. Show that, the sum of one number and the square of the other is at least Let represent the first number. Then k is the other number. Let S represent their sum. k. Equation to maimize or minimize. S k 0 k Since we need to find the minimum value of S(), we now differentiate set equal to zero, and solve for S 0 Endpoints The endpoints are =0 and =k Check S 0 k S k k S k Therefore k is the smallest sum, which occurs when one number is.

5 Find the dimensions of the rectangle of the largest area that can be inscribed in the circle of radius. Let A represent the area of rectangle Let represent the width of rectangle Let y represent the height of rectangle 8 y Relationships A, y y y 6 Pythagorean Theorem A(, y) format indicates A is in terms of two variables: and y We want A in terms of one variable so let s write y in terms of y 6 A Find the derivative and set to zero to find critical points A 6 6 A since 0 Now area is in terms one variable. Note the restriction of

6 Endpoints The endpoints are =0 and =8 Check A 0 0 A 3 A 8 0 Thus the largest are occurs when and y Therefore the dimensions of the rectangle are. Three identical kennels are to be constructed in an open area. If the length of the fencing used is 500 m, find the dimensions that will give a maimum area. w w w w l Let l represent the length of fencing on area Let w represent the width of fencing on area Let A represent the area Given: l w 500 Relationship: A Work: lw We want to write A in terms of only one variable. l

7 l w 500 w 500 l w 5 l A lw A l 5 l 0 50 l 5l l A l 5l A l 0 5 l 0 l 5 Check A A A The Maimum area is 650 and this occurs when l=5, therefore since l w 500, w =6.5. Conclusion: Therefore the dimensions are 5 m by 6.5 m Rogers Cable is laying optic cable in an area with underground utilities. Two subdivisions are located on opposite side of the Grand river, which is 300 m wide. The company has to connect points P and Q with cable, where Q is on the north bank 900m east of P. It cost $0/m to lay cable underground and $0/m to lay cable underwater. What is the least epensive way to lay the cable Q 300m y P

8 Let represent the point on the bank accross the river Let y represent the distance from P to Let C represent the total cost Given: y Required:, when C is a minimum Relationships: C=0y+0(900-) Total cost= (price per metre underwater) (distance underwater)+(price per metre over land) (distance over land) Work: Write C in terms of one variable C 0y Find derivative and set to zero, and solve for candidates. 0 C C Endpoints The endpoints are =0 and =900. Check C C C Conclusion: The least epensive way is to lay the cable underwater to a point 73. m east of P.

9 Find the turning points (local etrema) of the function f tan., where The derivative of f is f sec. To find the Turning Points, we solve f 0 ; that is sec Since solutions sec 0 cos cos cos is positive in the given interval, we discard the minus sigh, obtaining the and. In order to determine whether these values give turning points, we observe that the sign of f does not change on each of the sub-intervals.,, since signs of are the only values at which f f by evaluating it at one point of each sub-interval: f 3, This leads to the following conclusions : f 0, f 3 is zero. We can thus determine the during the interval during the interval during the interval the function is decreasing. the function is increasing the function is decreasing Thus is a local minimum, is a local maimum

10 The position of a particle as it moves horizontally is described by the equation s sin t cos t,0 t where s is the displacement in metres and t is the time in seconds. Find the maimum and minimum displacements. Recall: the maimum or minimum displacement (height) occurs when the velocity is zero. ds v dt cos t sin t Setting the velocity to zero and solving cos t sin t 0 sin t tan t cos t No need to solve for t, since we will be converting back to sine and cosines. Recall from special triangles that tan t O A O or A H O A 5 H 5 This technique is very useful. You can solve these questions without a calculator and obtain eact answers Therefore O A 3 sin t, cos t, t H 5 H 5 or O A sin t, cos t, t H 5 H 5 Because tan(t) had a negative value, t could be in quadrant or quadrant

11 The displacement is then (from calculus) s or s The end points also gives us s sin 0 cos 0 s sin cos Therefore the maimum displacement is 5 m 5 m and the minimum displacement is Find the maimum perimeter of a right angle triangle with hypotenuse 0 cm Label: Let base angle be, 0 Let represent the adjacent Let y represent the opposite Let P represent the perimeter y Given: hypotenuse=0 cm Required: P when P is a maimum Relationships: 0cos, y 0sin Work: P 0 y 0 0cos 0sin dp 0sin d 0cos Set derivative equal to 0 to find critical points

12 0sin 0cos 0 check P 0 0 sin cos tan P 0 0 P 0 Conclusion: The maimum perimeter is 0 0 cm A rectangular building is located on a corner. The city wants to build a sidewalk from one street to the other, just touching the back of a building. Find the angle in degrees of the sidewalk that minimizes the distance d along the walk. Label: d represents distance street to building d represents distance building to street d Given: d 7 Length of building 7 m Width of building is 6 m 6 Required: when d is a minimum Relationships: d d d Work: d d d d d d 6 7 sin cos 6 csccot 7sectan Set equal to zero and solve for angle

13 6csccot 7sectan 6cos 7sin sin cos 3 3 6cos 7sin tan Conclusion: The angle of the building is appro 53 degrees