1. Optimization: Cage Building Questions The Calculus Crusaders

10. Optimization (a) So the volume as a function of c will be: V(c) = c(4 - 2c)² where the length and width are equal to 4 - 2c and the height is equal to c .

11. Optimization (a) So are volume as a function of c will be: V(c) = c(4 - 2c)² Simplified, V(c) looks like so: V(c) = 16c – 16c² + 4c³

12. Optimization (a) So are volume as a function of c will be: V(c) = c(4 - 2c)² Simplified, V(c) looks like so: V(c) = 4c³ – 16c² + 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) Let’s see how the derivative of the volume function looks like: V’(c) = 16 – 32c + 12c²

18. Optimization (a) Let’s see how the derivative of the volume function looks like: V’(c) = 16 – 32c + 12c² 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 + 12c² 0 = 4 (4 – 8c – 3c²) 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 + 12c² 0 = 4 (4 – 8c – 3c²) 0 = 4 (3c -2) (c -2) Therefore there is a critical number @ c = ⅔ & c = 2

21. Optimization (a) Now we apply an interval analysis to find where the function is increasing or decreasing. ⅔ 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 can’t use the endpoints of the interval because that would make the amount of material at the corners being cut out so small that you can’t 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 at c = ⅔ by the First Derivative Test. ⅔ 2 V’ + + -

25. Optimization (a) Now we know that when the dimensions of the squares being cut out must be ⅔ m by ⅔ m to maximize the volume of the box.

26. Optimization (a) Now we know that when the dimensions of the squares being cut out must be ⅔ m by ⅔ m to maximize the volume of the box. The volume of the box is approximately 4.7407 m ³ which was found by solving for V(⅔).

45. At least the animals won’t get hurt anymore.