Troublesome Pets
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The Calculus Crusaders Optimization: Cage Building Questions
Optimization (a) Bench: We can’t afford to get these animals hurt again. We need to bring
them back home and put them in a box. Jamie won’t 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.
Optimization (a) Bench: Cool, the sheet is 4m by 4m, do you think we have enough? Zeph: I think so, let’s find out how we can maximize the volume of the box.
Bench: Okay, I’ll do the math.
Optimization (a) Let’s take a look at what we have:
Optimization (a) Let’s 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.
Optimization (a) Let’s 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?
Optimization (a) Here is the formula used to find the volume of the box: V=lwh
Optimization (a) Here is the formula used to find the volume of the box: V=lwh So what are the dimensions of the box?
Optimization (a) Here is the formula used to find the volume of the box: V=lwh So what are our dimensions of the box? When we cut the squares out, each side of the sheet was decreased by 2c.
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.
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
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.
Optimization (a) We could use grade 11 pre-calc to find the max or we could use “Calculus”.
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
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.
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.
Optimization (a) Let’s see how the derivative of the volume function looks like: V’(c) = 16 – 32c + 12c²
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.
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)
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
Optimization (a) Now we apply an interval analysis to find where the function is increasing or decreasing.
V’
⅔
2
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.
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.
Optimization (a) Now we apply an interval analysis to find where the function is increasing or decreasing.
V’
+
⅔
+ 2
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.
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.
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 m³ which was found by solving for V(⅔).
Optimization (b) Bench: Okay, we can build the box and I didn’t even use my calculator except for finding the volume. Zeph: Hey wait!!! Our animals can’t 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…
Optimization (b) The rate of Oxygen per hole is modeled by the function:
in cm³/hr/hole. While the rate of Carbon dioxide per hole is modeled by the function:
in cm³/hr/hole.
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.
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.
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.
Optimization (b) Since we are integrating the Oxygen or Carbon dioxide functions, it will be the integrand as shown below:
where R(h) is used to find the rate of oxygen and S(h) is used to find the rate of carbon dioxide. These functions represent the rate of gas entering the box, in cm ³/hr, depending on the number of holes drilled into the box
Optimization (b) Since we are integrating the Oxygen or Carbon dioxide functions, it will be the integrand as shown below:
where R(h) is used to find the rate of oxygen and S(h) is used to find the rate of carbon dioxide. These functions represent the rate of gas entering the box, in cm ³/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.
Optimization (b) I’ll 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.
Optimization (b) I’ll 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.
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) and S’(h) = C(h)
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) and S’(h) = C(h)
Now just set the functions equal to zero and find the roots. Remember the Extreme Value Theorem.
Optimization (b) R’(h) = O(h) and S’(h) = C(h) In the context of the question, the two endpoints of the interval can be included. In other words, 0 holes and h holes (what ever it is) must be included in the analysis.
Optimization (b) R’(h) has critical numbers at h = -2, 0, 11, 20 h = -2 can be rejected because it is not within the domain.
-
R’(h) 0
+ 11
20
Optimization (b) R’(h) has critical numbers at h = -2, 0, 11, 20 h = -2 can be rejected because it is not within the domain.
-
R’(h) 0
+ 11
20
h = 11 cannot be a maximum because it is decreasing to the left of 11 and increasing to the right. h = 0 cannot be a maximum because if 0 is used, you will be integrating from 0 to 0 which will produce 0 as an answer. So the only choice left would be h = 20.
Optimization (b) S’(h) has critical numbers at h = 0, 12, 20 There is one more critical number but it is rejected because it is not in the domain.
+
S’(h) 0
-
12
20
Optimization (b) S’(h) has critical numbers at h = 0, 12, 20 There is one more critical number but it is rejected because it is not in the domain.
+
S’(h) 0
-
12
20
h = 12 cannot be a minimum because S is increasing on the left of h = 12 and decreasing on the right, making it a maximum. h = 0 cannot be used because of the same reason as h = 0 in R’(h). Therefore the only choice left is h = 20.
Optimization (b) Let’s recap our answers. Therefore to obtain a maximum rate of oxygen and a minimum rate of carbon dioxide entering the box, there must be 20 holes drilled into the box.
Optimization Bench: There we go.You guys stay put, while we find the block of wood.
Zeph: Okay, let’s get out of here before Jamie comes and yells at us for caging up our animals.
At least the animals won’t get hurt anymore.
Optimization (a) Bench: We can’t afford to get these animals hurt again. We need to bring
them back home and put them in a box. Jamie won’t 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.
Optimization (a) Bench: Cool, the sheet is 4m by 4m, do you think we have enough? Zeph: I think so, let’s find out how we can maximize the volume of the box.
Bench: Okay, I’ll do the math.
Optimization (a) Let’s take a look at what we have:
Optimization (a) Let’s 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.
Optimization (a) Let’s 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?
Optimization (a) Here is the formula used to find the volume of the box: V=lwh
Optimization (a) Here is the formula used to find the volume of the box: V=lwh So what are the dimensions of the box?
Optimization (a) Here is the formula used to find the volume of the box: V=lwh So what are our dimensions of the box? When we cut the squares out, each side of the sheet was decreased by 2c.
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.
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
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.
Optimization (a) We could use grade 11 pre-calc to find the max or we could use “Calculus”.
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
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.
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.
Optimization (a) Let’s see how the derivative of the volume function looks like: V’(c) = 16 – 32c + 12c²
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.
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)
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
Optimization (a) Now we apply an interval analysis to find where the function is increasing or decreasing.
V’
⅔
2
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.
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.
Optimization (a) Now we apply an interval analysis to find where the function is increasing or decreasing.
V’
+
⅔
+ 2
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.
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.
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 m³ which was found by solving for V(⅔).
Optimization (b) Bench: Okay, we can build the box and I didn’t even use my calculator except for finding the volume. Zeph: Hey wait!!! Our animals can’t 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…
Optimization (b) The rate of Oxygen per hole is modeled by the function:
in cm³/hr/hole. While the rate of Carbon dioxide per hole is modeled by the function:
in cm³/hr/hole.
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.
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.
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.
Optimization (b) Since we are integrating the Oxygen or Carbon dioxide functions, it will be the integrand as shown below:
where R(h) is used to find the rate of oxygen and S(h) is used to find the rate of carbon dioxide. These functions represent the rate of gas entering the box, in cm ³/hr, depending on the number of holes drilled into the box
Optimization (b) Since we are integrating the Oxygen or Carbon dioxide functions, it will be the integrand as shown below:
where R(h) is used to find the rate of oxygen and S(h) is used to find the rate of carbon dioxide. These functions represent the rate of gas entering the box, in cm ³/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.
Optimization (b) I’ll 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.
Optimization (b) I’ll 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.
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) and S’(h) = C(h)
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) and S’(h) = C(h)
Now just set the functions equal to zero and find the roots. Remember the Extreme Value Theorem.
Optimization (b) R’(h) = O(h) and S’(h) = C(h) In the context of the question, the two endpoints of the interval can be included. In other words, 0 holes and h holes (what ever it is) must be included in the analysis.
Optimization (b) R’(h) has critical numbers at h = -2, 0, 11, 20 h = -2 can be rejected because it is not within the domain.
-
R’(h) 0
+ 11
20
Optimization (b) R’(h) has critical numbers at h = -2, 0, 11, 20 h = -2 can be rejected because it is not within the domain.
-
R’(h) 0
+ 11
20
h = 11 cannot be a maximum because it is decreasing to the left of 11 and increasing to the right. h = 0 cannot be a maximum because if 0 is used, you will be integrating from 0 to 0 which will produce 0 as an answer. So the only choice left would be h = 20.
Optimization (b) S’(h) has critical numbers at h = 0, 12, 20 There is one more critical number but it is rejected because it is not in the domain.
+
S’(h) 0
-
12
20
Optimization (b) S’(h) has critical numbers at h = 0, 12, 20 There is one more critical number but it is rejected because it is not in the domain.
+
S’(h) 0
-
12
20
h = 12 cannot be a minimum because S is increasing on the left of h = 12 and decreasing on the right, making it a maximum. h = 0 cannot be used because of the same reason as h = 0 in R’(h). Therefore the only choice left is h = 20.
Optimization (b) Let’s recap our answers. Therefore to obtain a maximum rate of oxygen and a minimum rate of carbon dioxide entering the box, there must be 20 holes drilled into the box.
Optimization Bench: There we go.You guys stay put, while we find the block of wood.
Zeph: Okay, let’s get out of here before Jamie comes and yells at us for caging up our animals.
At least the animals won’t get hurt anymore.
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