The Exciting World of Optimization

1. The Exciting World ofOptimization Emma Sullivan

2. Optimization • Optimization is the procedure or procedures used to make a system as effective or functional as possible. • In this case, we will be using optimization to maximize area and volume.

3. A rectangular patch of land is to be enclosed by 60 ft. of fencing for a doggie play pen. Find the length and width that will give the maximum area. • The first step is choose which formula fits the problem. • Drawing a picture is helpful. • For this example, the formula for the perimeter of a rectangle is needed • A = 2L + 2w

4. A rectangular patch of land is to be enclosed by 60 ft. of fencing for a doggie play pen. Find the length and width that will give the maximum area. • Next, plug in the information that the problem gives you. • You know that the maximum fencing is 60, so 60 = 2L + 2w • Then solve for one of the variables. 2L = 60 – 2w L = 30 – w

5. A rectangular patch of land is to be enclosed by 60 ft. of fencing for a doggie play pen. Find the length and width that will give the maximum area. • Next, plug your L value back into the original perimeter formula. 2(30 – w) + 2w = 60 • Now since we are maximizing area, we need to consider the area formula. A= wL

6. A rectangular patch of land is to be enclosed by 60 ft. of fencing for a doggie play pen. Find the length and width that will give the maximum area. • Using your values from 2(30 – w) + 2w = 60 Plug them into the area formula A = 2(30 – w)(2w) • Simplify A = (60 – 2w)(2w) A = 120w – 4w2

7. A rectangular patch of land is to be enclosed by 60 ft. of fencing for a doggie play pen. Find the length and width that will give the maximum area. • Now, take the derivative of the equation A’ = 120 – 8w • Set equal to zero and solve for w 0 = 120 – 8w 8w = 120 w = 15

8. A rectangular patch of land is to be enclosed by 60 ft. of fencing for a doggie play pen. Find the length and width that will give the maximum area. • Finally, plug the value of w back into the original perimeter equation to find the value of L 2L + 2(15) = 60 2L + 30 = 60 2L = 30 L = 15 • The length and width that will give the maximum area are 15 ft X 15 ft.

9. An origami box is to be made by cutting squares from the corners of a 20 x 20 cm square piece of paper and bending up the sides. How long should the side of the square be so that the box has a maximum volume? • Let each side of the soon-to-be cut out squares be represented by x • Write the length, width, and height of the paper in terms of x w = 20 – x L = 20 – x H = x

10. An origami box is to be made by cutting squares from the corners of a 20 x 20 cm square piece of paper and bending up the sides. How long should the side of the square be so that the box has a maximum volume? • Plug your new length, width, and height into the formula for volume V = Lwh V = (20 – x)2 x • Simplify V = (4x2 – 80x + 400)x V = 4x – 80x2 + 400x

11. An origami box is to be made by cutting squares from the corners of a 20 x 20 cm square piece of paper and bending up the sides. How long should the side of the square be so that the box has a maximum volume? • Next, take the derivative V’ = 12x2 – 160x + 400 • Simplify 4(3x2 – 30x + 100) 4(3x – 10)(x – 10)

12. An origami box is to be made by cutting squares from the corners of a 20 x 20 cm square piece of paper and bending up the sides. How long should the side of the square be so that the box has a maximum volume? • Set derivative equal to zero and solve for x x = 10/3 and x = 10 • Since the sides of the paper are only 20 cm, the squares cannot be 10 • Therefore, the answer is 10/3 cm.

13. You have now learned optimization! Give yourself a big pat on the back!