1. Real Life Quadratic Equations Maximization Problems Optimization Problems Module 10 Lesson 4:

2. Table of Contents  Slides 3-12: Maximization and Optimization (Height, Min/Max) Examples

3. The Height Formula

4. h= ending height g = gravity constant (32 if feet, 9.8 if meters) v 0 = initial velocity h 0 = initial height t = time If the problem is dealing with feet If the problem is dealing with meters

5. Find the maximum height of an object shot upward from the earth’s surface with an initial velocity of 106 ft/sec. Since the problem deals with feet, use h= ending height g = gravity constant v 0 = initial velocity h 0 = initial height t = time h= ending height g = (taken care of by using this equation) v 0 = GIVEN: 106 ft/sec h 0 = 0, since we are starting on the ground t = time Plug in what we know so far: Continued on next slide

6. Next we need to graph this. Changing t’s to x’s, we plug this in: You will need to ‘Zoom Fit’ and maybe ‘Zoom In’ or ‘Zoom Out” to see the graph correctly: You need to find the Maximum of the graph, (x, y), the x value (time), the y value (height) will be used for your answer. Your solution is the ‘y-value’ of the maximum. The maximum height is 175.56 feet Note: It takes x, 3.31 seconds, to reach the height, y, of 175.56 ft.

7. A ball is thrown vertically upward at an initial speed of 48 ft/sec from the top of a building 2000 feet high. How high does the ball go? Since the problem deals with feet, use h= ending height g = gravity constant v 0 = initial velocity h 0 = initial height t = time Plug in what we know so far: Next we need to graph this. Zoom to see the graph correctly. Find the Maximum of the graph, (x, y) Your solution is the ‘y-value’ of that point. Find the Maximum of the graph, (x, y) Your solution is the ‘y-value’ of that point. (1.5, 2036) So the maximum height of the ball is 2036 feet.

8. A bouquet is thrown straight up with an initial velocity of 56 ft/sec. The height of the ball in seconds after it is thrown is given by what equation? (We are assuming initial height is 0, unless otherwise specified) What is the maximum height of the ball? What is the height of the ball after 1 second? After how many seconds will the ball return to the ground? Explanation: To find the max height you should go to 2 nd trace, value, then x = 1 Explanation: to find when an object hits the ground you are looking for the x-intercept (or zero) on the right hand side of your graph. You can find this by using the following steps: 1. Go back to y= and put 0 in the y 2 equation. 2. Hit graph 3. Go to 2 nd trace, intersection (5), move your cursor over to the right hand x – intercept, then hit enter 3 times. You are looking for the x value.

9. A ball is dropped from the top of a 500 ft. building. How long does it take the ball to reach the ground? Initially, the velocity is 0 because it is dropped – so it was still just before it was let go. The building is 500 feet tall, so this is the initial height. Notice this graph is different. You must look at the y- intercept, (0, 500). This represents the ball just before it is released. The downward arc is the fall of the ball (decreasing height, y) over increasing time (x). When the graph hits the x-axis on the positive side, the graph is representing the time the ball hits the ground. Therefore it takes 5.59 seconds for the ball to hit the ground. y x (time) (height) (0,500) (1, 484) (2, 436) (3, 356) (4, 244) (5, 100) (5.59, 0) (x,y) (time, height)

10. A ball is thrown from the top of a tower at a velocity of 10 m/sec. The ball hit the ground at the base of the tower 4 seconds later. How high is the tower? Initially, the velocity is 10 meters per second. Because it hits the ground in 4 seconds, t = 4. Also, since the final height is 0, we make h = 0. ***h 0 is initial height, not ending height, so we are looking for this value. Since the problem deals with meters, use This means the tower that the ball was thrown from is 38.4 meters tall.

11. Step 6: x = 18, plug into x + y = 36, thus, y = 18 also (18,18). Find two positive numbers whose sum is 36 and whose product is a maximum. Step 1: Set up two equations: x + y = 36 xy = ? Step 2: Solve the first equation for y. y=-x + 36 Step 3: Plug the equation in Step 2 into xy = ? x(-x + 36) = ? Step 4: Put the equation in Step 3 into your graphing calculator. Y = -x 2 + 36x ************************** Zoom Fit to graph Step 5: Find the maximum of the graph. 2 nd Calc: Maximum Move to the left of the maximum point, press ENTER Move to the right of the maximum point, press ENTER

12. A rectangle has a perimeter of 40 meters. Find the dimensions of the rectangle with the maximum area. Recall: Perimeter = 2L + 2W Recall: Perimeter = 2L + 2W L W Step 1: Set up two equations: 2x + 2y = 40 xy = ? Step 2: Solve the first equation for y. 2y= - 2x + 40 y = - x + 20 Step 3: Plug the equation in Step 2 into xy = ? x( - x + 20) = ? Step 4: Put the equation in Step 3 into your graphing calculator. y = -x 2 +20x ***********Zoom Fit to graph Step 5: Find the maximum of the graph. Step 6: x = 10, plug into 2x + 2y = 40, thus, y = 10 also (10, 10). L =10 and W =10