Optimization Which point on the line y = 1 2x. is closest to the origin? MATH 1380 Lecture 18 1 of 15 Ronald Brent 2018 All rights reserved.

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1 Optimization Which point on the line y = 1 is closest to the origin? y MATH 1380 Lecture 18 1 of 15 Ronald Brent 018 All rights reserved.

2 Recall the distance between a point (, y) and (0, 0) is =. d + y If we require the point to be on the line i.e. y = 1, then the distance is d ( ) = + (1 ) = To find the absolute minimum of this function we first find critical points which equals 0 at d ( ) = =. The point is 1,. And the distance is d ( /5) =. 5 MATH 1380 Lecture 18 of 15 Ronald Brent 018 All rights reserved.

3 The method for solving optimization problems is as follows*: 1. Understand the Problem. What is being asked? Can you envision the problem? What is the numerical information being given? What are the units?. Build a Mathematical Model This will involve drawing pictures, labeling variables and using either algebraic, geometric, or trigonometric laws to write down equations relating the variables. a. One equation will be the function you wish to maimize or minimize. This is called the objective function. It will depend upon one or more variables. b. One equation will relate the variables in the problem. This is called the constraint equation. c. Using the constraint and the objective, many times the objective function can be reduced to a function of only one variable. MATH 1380 Lecture 18 3 of 15 Ronald Brent 018 All rights reserved.

4 3. What is the domain of the objective function? This needs to be determined from all the given data and the constraints. 4. Determine Critical Numbers. We know how to do this 5. Solve the Mathematical model. Determine all the variables leading to optimized solutions, use them in various formula to determine actual requested solution. 6. Interpret the Solution. What does the mathematical solution mean in terms of the originally stated problem? Write it down in a coherent sentence! MATH 1380 Lecture 18 4 of 15 Ronald Brent 018 All rights reserved.

5 Maimizing area given perimeter: Suppose that the perimeter of a rectangle is 0 feet. What are the dimensions of the rectangle with the greatest area? 1) Pretty self-eplanatory here. This is a geometry problem concerning area and length (perimeter.) Units are in feet. ) y A perimeter of 0 feet translates to + y = 0, so + y = 10. Maimize the area, A = y (objective) subject to + y = 10 (constraint.) + y = 10 implies y = 10, and so A( ) = (10 ). MATH 1380 Lecture 18 5 of 15 Ronald Brent 018 All rights reserved.

6 3) The perimeter constraint restricts between 0 and 10. 4) Since A ( ) = 10, the critical point is = 5. 5) With A ( ) =, the second derivative test insures that A ( 5) = 5 is a maimum. Endpoints yield A ( 0) = A(10) = 0 the minimum area. Desired solution is = 5, which means y = 5. 6) The dimensions of the rectangle with perimeter 0 feet with the largest area is a square of side length 5 feet. Definition: A constrained optimization problem is one where you are given an objective function to maimize or minimize subject to a constraint function. In the last eample we solved the constraint function for one of the variable in terms of the other, and substituted the result into the objective function. We then used typical methods to find absolute maimum values for the resulting equation. This is not always possible. MATH 1380 Lecture 18 6 of 15 Ronald Brent 018 All rights reserved.

7 Eample: A piece of wire 10 cm long is cut into pieces. One piece is shaped into a circle and the other into a square. Find the dimensions of the shapes that maimize and minimize the area. 10 cm y r First off, + y = 10. Net, we need the radius of the circle to compute its area. What we have is, the circumference. If the piece meters long is bent into the circle, then = π r, so r = and the area of the circle is π circle A = π r = π. π MATH 1380 Lecture 18 7 of 15 Ronald Brent 018 All rights reserved.

8 The remaining piece, length y is bent into the square, so it has side length 4 y and so the area of the square is Since + y = 10, A square 4 = y Which gives Total area is A square y = = 4 10 A( ) = A + A = π + circle square π 4 Simplifying the area function, we obtain MATH 1380 Lecture 18 8 of 15 Ronald Brent 018 All rights reserved.

9 A( ) = + 4π We wish to optimize this function over the interval [0, 10]. Since π 10π A ( ) = + =, π 8 8π the critical point is so that 10π 1 1 =. Since A ( ) = + > π π 8, this yields a minimum The minimum area becomes 5π A = + = cm (4 + π ) (4 + π ) 4 + π 5 Note: The maimum area occurs at one of the endpoints. A ( 0) = cm, and 4 5 A ( 10) = cm. So, the maimum occurs when only a circle is built. π MATH 1380 Lecture 18 9 of 15 Ronald Brent 018 All rights reserved.

10 Eample: Minimize the surface area of a can with 1 cubic foot volume. Let r be the radius of the can, and h the height. r The volume is V = π r h The surface area is S = π rh + ( π r ) Given the volume constraint 0 < r < h For V = 1, we have 1 π r 1 = h or h = π r This gives a surface area of S = + ( π r ) r Hence, S = 4π r r + 1. Critical points occur at r = 0, and r = ft, since r = 0 3 π isn t in the domain of the problem, we don t consider it. 1 When r = ft, h = ft. The surface area is S = 3 π sq. ft. 3 π π MATH 1380 Lecture of 15 Ronald Brent 018 All rights reserved.

11 Eample: Building an open topped bo V = (1 )(1 ) = So V = = = 0 ( 6)( ) = 0 0 Critical points =, 6. V()=18 and V(6)=0 Ma is at = with volume 18 MATH 1380 Lecture of 15 Ronald Brent 018 All rights reserved.

12 MATH 1380 Lecture 18 1 of 15 Ronald Brent 018 All rights reserved.

13 Eample: Your dreams of becoming a hamster breeder have finally come true!!! You are constructing a set of rectangular pens in which to house your furry friends. The overall area you are working with is 60 square feet, and you want to divide the area up into si pens of equal size as shown below. The cost of the outside fencing is $10 a foot. The inside fencing costs $5 a foot. You wish to minimize the cost of the fencing. a) Labeling variables, write down a constrained optimization problem that describes this problem. b) Using any method learned in this course, find the eact dimensions of each pen that will minimize the cost of the breeding ground. What is the total cost? MATH 1380 Lecture of 15 Ronald Brent 018 All rights reserved.

14 The cost of the outside fencing is $10 a foot. The inside fencing costs $5 a foot. You wish to minimize the cost of the fencing. y 10 sq.ft. a) Let be the width of each individual pen, and y be the length as shown above. Since the total area is 60 sq. ft., each individual pen will have an area of 10 sq. ft. The constraint is y = 10. The objective function is the cost. Eamining the fencing above, there is 5 y feet of interior fencing, and y + 1 feet of eterior fencing. So the total cost is C = $ 5 (5y) + $10 (y + 1), or C = 45 y MATH 1380 Lecture of 15 Ronald Brent 018 All rights reserved.

15 The constrained optimization problem is: Minimize to the constraint y = 10. C = 45 y + 10 subject b) Solving y = 10 for y gives 10 y =. Substituting this into C gives 450 C = + 10, as the function to minimize over ( 0, ). 450 C = + 10, Setting C = 0 gives the equation: = or = = so = =,so critical points are = 0, and =, and so y = =. 3 The cost is C = 45y + 10 = $ MATH 1380 Lecture of 15 Ronald Brent 018 All rights reserved.