1. Calculus I (MAT 145) Dr. Day Monday April 8, 2019
Chapter 4: Using All Your Derivative Knowledge!
Absolute and Relative Extremes
What is a “critical number?”
Increasing and Decreasing Behavior of Functions
Connecting f and f’
Concavity of Functions: A function’s curvature
Connecting f and f”
Graphing a Function: Putting it All together!
Max-Mins Problems: Determine Solutions for Contextual Situations
Finally What if We Reverse the Derivative Process?
Monday, April 8, 2019
MAT 145

2. Optimization What does it mean to optimize?
What are examples in which you might want to determine an optimum solution?
What should you consider when looking for an optimum solution?
Monday, April 8, 2019
MAT 145

3. Optimization
Two numbers have a sum of48. If the product of the two numbers is to be as large as possible, what are the two numbers?
Monday, April 8, 2019
MAT 145

4. Optimization
An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks on the south bank of the river 6 km east of the refinery. The cost of laying pipe is $400,000 per km over land and $800,000 per km under the river to the tanks. How should the oil company lay the pipe in order to minimize the cost? What is the minimum cost?
Monday, April 8, 2019
MAT 145

5. Optimization
Jane is 2 miles offshore in a boat and wants to reach a village, on the coast, that is 6 miles down a straight shoreline from the point nearest her boat. Jane can row 2 mph and she can walk 5 mph. Where should she land her boat in order to reach the village in the least amount of time?
Monday, April 8, 2019
MAT 145

6. Open top box
A square piece of cardboard 1 foot on each side is to be cut and folded into a box with no top. To accomplish this, congruent squares will be cut from each corner of the cardboard. Determine the dimensions of the box that will enclose the greatest volume.
Monday, April 8, 2019
MAT 145

7. Optimization: An example
A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. She needs no fence along the river. What are the dimensions of the field that has the largest area?
What are we trying to optimize here and in what way (greatest or least)?
Which variables are involved?
What constraints should be taken into account?
What initial guesses do you have about a solution?
Monday, April 8, 2019
MAT 145

8. Optimization: Guess & check
A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. She needs no fence along the river. What are the dimensions of the field that has the largest area?
What are options? Guess and check to fit constraints. Are these all the options?
Monday, April 8, 2019
MAT 145

9. Optimization: Calculus strategy
A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. She needs no fence along the river. What are the dimensions of the field that has the largest area?
Let x and y be the length and width of the rectangular field (in feet). Then we express A, area of rectangular field (in sq. feet) in terms of x and y: A = xy
Optimization Equation: A=xy
Constraint: 2x + y = 2400
Combine: A = x(2400 – 2x) = 2400x – 2x2
Domain: x is within [0, 1200]
Now, find absolute maximum of A.
(Just like we did when graphing.)
Monday, April 8, 2019
MAT 145

10. Optimize
Equation to maximize: A (x) = 2400x – 2x2 , A represents area of field in sq. ft; x represents width of field, as shown in the diagram.
Domain: 0  x  1200; Thus, endpoints to check are x=0, x=1200.
Find derivative and critical numbers (locations at which derivative is zero or undefined):
A (x) = 2400 – 4x
2400 – 4x = 0
x = 600
The absolute maximum value of A must occur either at this critical number or at an endpoint of the interval.
Since A(0) = 0, A(600) = 720,000, and A(1200) = 0, the Closed Interval Method gives the maximum value as A(600) = 720,000.
Dimensions of field with greatest area: 600 ft by 1200 ft.
Note: A’ function is a polynomial; so A’ is defined everywhere on its domain.
Monday, April 8, 2019
MAT 145

11. Steps for Optimizing
Understand the problem: What is given? What is requested? Are there constraints?
Diagram and variables: If appropriate for the context draw a diagram and label with variables. Define variables (by saying what the letters you are using represent, including units) and determine a reasonable span of values for those variables in the context of the problem. This will help you identify the domain of the independent variable.
Determine quantity to be optimized and write an equation: Read the problem to determine what you have been asked to optimize. Express the quantity to be optimized (one of your variables), as a function of the other variable(s) in the problem.
Identify constraint equation, if appropriate: If there are several variables, one may be constrained to be a function of the others. Write an equation that relates those variables. Combine with other equation previously defined, if appropriate.
Use the closed interval method to identify the absolute maximum or minimum
Answer the question!
Monday, April 8, 2019
MAT 145

12. Bendable Wire
A bendable wire measures 1000 cm in length. Write a function, call it A(x), to represent the sum of the areas of a circle and a square that result when that wire is cut at one point and the resulting two pieces of wire are used to create those shapes. Determine how to cut the wire so that the sum of the areas of the circle and the wire are a maximum.
Monday, April 8, 2019
MAT 145

13. Bendable Wire
Monday, April 8, 2019
MAT 145

14. Bendable Wire
Note: The A’ function is a polynomial; so A’ is defined everywhere on its domain. Thus, the only critical numbers occur when A’ = 0.
A(4000/(π+4)) ≈35,006, A(0)≈79,577, A(1000)=62,500. Thus, max area occurs when x=0 and entire wire is used to make a circle.
Monday, April 8, 2019
MAT 145

15. Solve the follow optimization problem
Solve the follow optimization problem. Show complete evidence and calculus justification. Include a drawing or a graph to represent the situation. Include evidence that shows you have considered any domain restrictions.
____ 1 pt: labeled sketch, drawing, graph;
____ 1 pt: variables identified/described;
____ 1 pt: domain of independent variable;
____ 1 pt: statement of optimizing function;
____ 1 pt: constraint;
____ 1 pt: calculus evidence;
____ 1 pt: justify optimum;
____ 1 pt: consider all possibilities for critical points;
____ 1 pt: correct solution, units labeled
Monday, April 8, 2019
MAT 145

16. Wednesday, Oct 31, 2018
MAT 145
Recovering Functions

17. Recovering Functions: Solutions p1 Solutions p2
Wednesday, Oct 31, 2018
MAT 145

18. Position, Velocity, Acceleration
An object is moving in a positive direction when ….
An object is moving in a negative direction when ….
An object speeds up when ….
An object slows down when ….
An object changes directions when ….
The average velocity over a time interval is found by ….
The instantaneous velocity at a specific point in time is found by ….
The net change in position over a time interval is found by ….
The total distance traveled over a time interval is found by ….
Wednesday, Oct 31, 2018
MAT 145

19. Position, Velocity, Acceleration
An object is moving in a positive direction when v(t) > 0.
An object is moving in a negative direction when v(t) < 0.
An object speeds up when v(t) and a(t) share same sign.
An object slows down when v(t) and a(t) have opposite signs.
An object changes directions when v(t) = 0 and v(t) changes sign.
The average velocity over a time interval is found by comparing net change in position to length of time interval (SLOPE!).
The instantaneous velocity at a specific point in time is found by calculating v(t) for the specified point in time.
The net change in position over a time interval is found by calculating the difference in the positions at the start and end of the interval.
The total distance traveled over a time interval is found by first determining the times when the object changes direction, then calculating the displacement for each time interval when no direction change occurs, and then summing these displacements.
Wednesday, Oct 31, 2018
MAT 145

20. Position, velocity, acceleration
Velocity: rate of change of position
Acceleration: rate of change of velocity
Velocity and acceleration are signed numbers. Sign of velocity (pos./neg.) indicates direction of motion (right/left or up/down)
When velocity and acceleration have the same sign (both pos. or both neg.), then object is speeding up. This is because object is accelerating the same direction that the object is moving.
When velocity and acceleration have opposite sign (one positive and one negative), then object is slowing down. This is because object is accelerating the opposite direction that the object is moving.
Wednesday, Oct 31, 2018
MAT 145