# troublesome pets

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- 1. Optimization: Cage Building Questions The Calculus Crusaders

2. Optimization (a)

- Bench:

- We cant afford to get these animals hurt again. We need to bring them back home and put them in a box. Jamie wont know.

- Zeph:

- Hey, look! There is a guy over there selling a sheet of plastic that we can use to make a box. We can build a box by cutting squares from the corners.

3. Optimization (a)

- Bench:

- Cool, the sheet is 4m by 4m, do you think we have enough?

- Zeph:

- I think so, lets find out how we can maximize the volume of the box.

- Bench:

- Okay, Ill do the math.

4. Optimization (a)

- Lets take a look at what we have:

5. Optimization (a)

- Lets take a look at what we have:

- So we must make the box by cutting squares from the corners and folding up the remaining sides.

6. Optimization (a)

- Lets take a look at what we have:

- So we must make the box by cutting squares from the corners and folding up the remaining sides.

- What should be the dimensions of the squares to maximize the volume of the box? What is the volume of the box?

7. Optimization (a)

- Here is the formula used to find the volume of the box:V = l w h

8. Optimization (a)

- Here is the formula used to find the volume of the box:V = l w h

- So what are the dimensions of the box?

9. Optimization (a)

- Here is the formula used to find the volume of the box:V = l w h

- So what are our dimensions of the box?

- When we cut the squares out, each side of the sheet was decreased by 2c.

10. Optimization (a) So the volume as a function of c will be: V(c) = c(4 - 2c) where the length and width are equal to4 - 2cand the height is equal toc . 11. Optimization (a) So are volume as a function of c will be: V(c) = c(4 - 2c) Simplified, V(c) looks like so: V(c) = 16c 16c + 4c 12. Optimization (a) So are volume as a function of c will be: V(c) = c(4 - 2c) Simplified, V(c) looks like so: V(c) = 4c 16c + 16c This is the volume function. In order to maximize the volume, we need to find the maximum value of the function. 13. Optimization (a) We could use grade 11 pre-calc to find the max or we could use Calculus . 14. Optimization (a) We could use grade 11 pre-calc to find the max or we could use Calculus To use Calculus, we can use the First Derivative Test. Finding the max using the test requires finding where the volume function is increasing and then decreasing 15. Optimization (a) So we find the first derivative of the function. This can be done by using the power rule for each term of the polynomial. 16. Optimization (a) So we find the first derivative of the function. This can be done by using the power rule for each term of the polynomial. The power rule says that the derivative of any variable to an exponent can be found by multiplying the term by the exponent and decrease the exponent by 1. 17. Optimization (a) Lets see how the derivative of the volume function looks like: V(c) = 16 32c + 12c 18. Optimization (a) Lets see how the derivative of the volume function looks like: V(c) = 16 32c + 12c Now we need to find the zeroes of the derivative, because a function has a max or min when its derivative has a zero/root or is undefined. The x-coordinate of the root is called a critical number. 19. Optimization (a) Set the function equal to zero and solve for the variable, in this case,c. 0 = 16 32c + 12c 0 = 4 (4 8c 3c) 0 = 4 (3c -2) (c -2) 20. Optimization (a) Set the function equal to zero and solve for the variable, in this case,c. 0 = 16 32c + 12c 0 = 4 (4 8c 3c) 0 = 4 (3c -2) (c -2) Therefore there is a critical number @ c =& c = 2 21. Optimization (a) Now we apply an interval analysis to find where the function is increasing or decreasing. 2 V 22. Optimization (a) Now we apply an interval analysis to find where the function is increasing or decreasing. You may remember the Extreme Value Theorem which states that the endpoints of the intervals are candidates for maximums and minimums. 23. Optimization (a) Now we apply an interval analysis to find where the function is increasing or decreasing. You may remember the Extreme Value Theorem which states that the endpoints of the intervals are candidates for maximums and minimums. But we cant use the endpoints of the interval because that would make the amount of material at the corners being cut out so small that you cant fold it or too large that there is nothing to fold. 24. Optimization (a) Now we apply an interval analysis to find where the function is increasing or decreasing. Now you can sort of visualize where the volume is increasing or decreasing from this analysis. We want to know where the function is increasing then decreasing and in this case it is atc =by the First Derivative Test. 2 V + + - 25. Optimization (a) Now we know that when the dimensions of the squares being cut out must bem bym to maximize the volume of the box. 26. Optimization (a) Now we know that when the dimensions of the squares being cut out must bem by m to maximize the volume of the box. The volume of the box is approximately 4.7407 m which was found by solving for V(). 27. Optimization (b)

- Bench:

- Okay, we can build the box and I didnt even use my calculator except for finding the volume.

- Zeph:

- Hey wait!!! Our animals cant breathe! We need to make holes. How many holes do we need to maximize the rate of Oxygen entering the box and minimize the rate of Carbon dioxide entering the box? There is a maximum of 20 holes.

- Bench:

- Since this box was made using math, the rate of Oxygen and Carbon dioxide entering the box must be mathematic as well.

- Zeph:

- Okay. Whatever you say

28. Optimization (b)

- The rate of Oxygen per hole is modeled by the function:

- in cm / hr/hole. While the rate of Carbon dioxide per hole is modeled by the function:

- incm / hr/hole.

29. Optimization (b)

- We need to maximize the rate and we were given the rate per hole. So if we integrate the function, we will find the total rate of Oxygen or Carbon dioxide.

30. Optimization (b)

- We need to maximize the rate and we were given the rate per hole. So if we integrate the function, we will find the total rate of Oxygen or Carbon dioxide.

- But we do not know how many holes are needed, which means we do not know how far we need to integrate to from zero.

31. Optimization (b)

- We need to maximize the rate and we were given the rate per hole. So if we integrate the function, we will find the total rate of Oxygen or Carbon dioxide.

- But we do not know how many holes are needed, which means we do not know how far we need to integrate to from zero.

- So we need to create a function that we can optimize that involves an integration from zero to a certain number of holes.

32. Optimization (b)

- Since we are integrating the Oxygen or Carbon dioxide functions, it will be the integrand as shown below:

- whereR(h)is used to find the rate of oxygen andS(h)is used to find the rate of carbon dioxide. These functions represent the rate of gas entering the box, in cm /hr, depending on the number of holes drilled into the box

33. Optimization (b)

- Since we are integrating the Oxygen or Carbon dioxide functions, it will be the integrand as shown below:

- whereR(h)is used to find the rate of oxygen andS(h)is used to find the rate of carbon dioxide. These functions represent the rate of gas entering the box, in cm/hr,depending on the number of holes drilled into the box

- These functions are called accumulation functions because you are accumulation areas. The h represents the number of holes. So we need to find what value of h will maximize the rate.

34. Optimization (b)

- Ill start with maximizing the rate of oxygen. As shown in making the box, to find the maxima or minima of a function, you can use the First Derivative Test. As a recap, a maximum is found where the derivative goes from positive to negative.

35. Optimization (b)

- Ill start with maximizing the rate of oxygen. As shown in making the box, to find the maxima or minima of a function, you can use the First Derivative Test. As a recap, a maximum is found where the derivative goes from positive to negative.

- The derivative of an accumulation function represents the second part of the Fundamental Theorem of Calculus, which states that the derivative of an integral is equal to the integrand. In other words, a derivative and an integral are inverses of each other.

36. Optimization (b)

- As stated earlier, the derivative of the accumulation functions is the integrand. Just switch the independent variable with the variable of the integral because accumulation functions are composites of functions which means an application of the Chain Rule is required.

- R(h) = O(h)andS(h) = C(h)

37. Optimization (b)

- As stated earlier, the derivative of the accumulation functions is the integrand. Just switch the independent variable with

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