Optimization: Follow the steps- The first thing you’re going to do is find the “given”, or the information the problem provides for you. Then you’re going to find a primary equation using the information they gave you. Next you’re going to find a secondary equation which is basically the given equation solved for either x or y. Then you are going to plug back into the primary equation. Then you’re going to take the derivative of the answer you just got from plugging in, and solve it for x. After doing that, you set up intervals to find if the function is increasing or decreasing. **Hint: if you make y 0 in the secondary equation it will tell you what the endpoint of your interval should be, which makes it easier to pick a point to plug into the derivative. If the interval isn’t going from –infinity, infinity then you’re going to have to check intervals by plugging all interval points into the given equation, to find the highest point. The highest point will be the answers.
Example: Find 2 non-negative numbers whose sum is 9 and so that the product of 1 number and the square of the other is a maximum. 1. Given: x+y=9 2. Primary: P=xy^2 3. Secondary: x+y=9 Solved: y=9-x *Making y 0, x=9 4. Plug in: P=x(9-x)^2;=x^3-18x^2+81x 5. Maximize: 3x^2-36x+81; x=9,x=3 (0,3)u(3,9);f’(1)=+,f’(4)=- Check: X=0, y=9 P=(0)(9)=0;x=3,y=6 P=(3)(6)=18; x=9,y=0 P=(9)(0) X=3, y=6 yields the highest product
**How do you do optimization? -Optimization basically means "maximizing" or "minimizing" so that means you're going to use the First Derivative Test. But before you can do that, you have to set up your own equations based off of the word problem. So the first thing you do is write down the formulas they give you and then determine which one is your primary formula and which one is the secondary formula. (*primary equation is the one that's closest to the words "maximum"/"minimum") Then you take your secondary equation and solve it for either x or y...I always solve for y, just to keep things simple. Then you plug your y into your primary equation and simplify it using algebra. Once you do that you take the derivative of the equation and solve for your x-values by setting it equal to zero. Then you set up your intervals, and lastly, you check your endpoints to see which one has the max/min value. Here's an example:
Ex) Determine the point(s) on y=x^2+1 that are the closest to (0,2). *First, let's pull apart the directions and try to decipher what they mean *When they say "closest" that means you're going to use the distance formula and it means "minimizing"..so you're looking for a min and you're going to use the first derivative test *But first we need to plug into the distance formula. So asuming that the point you were given in the directions is x1 and y1, y2 must be the equation you were given in the directions. So when you plug in you should get this: (x-0)^2+((x^2+1)-2)^2 ..*under a squareroot of course *Now you're going to simplify that^ so you should get this under the square root: x^4-x^2+1 *Now you can change it to (x^4-x^2+1)^1/2 because you have to take the derivative of it *So deriving it you get (4x^3-2x)/2sqrt(x^4-x^2+1) *Now take the numerator and set it equal to zero to find your x-values. 4x^3-2x=0 2x(2x^2-1)=0 x=0 x=+/-sqrt2/2 *Now you set up your intervals like this: (-inf,-sqrt2/2)u(-sqrt2/2,0)u(0,sqrt2/2)u(sqrt2/2,inf) *For the 1st interval you can plug in -1 and you should get negative, for the second you can plug in -.5 and you should get positve, for the third you can plug in .5 and you should get positive again, then for the last interval you can plug in 1 and you should get negative *So that means at -sqrt2/2 you have a min, nothing at 0, and a max at sqrt2/2 *So since -sqrt2/2 is a min, you can plug that into the original equation to get the y-value for your point *And you should get 3/2, so your final answer is: (-sqrt2/2,3/2)
Okay, optimization is when you are maximizing or minimizing something. The first thing you need to do is read the problem and figure out if you’re minimizing or maximizing. Then, you need to figure out your two equations: primary and secondary equation. Now, your primary equation is the equation used for max/minimizing. And your secondary equation is the equation given to you in the problem. After this, you need to solve your secondary equation for one of the variables. Then, plug your solved secondary equation into your primary equation and simplify. Finally, we begin max/minimizing. First of all, you take the derivative of the equation you just simplified. Then set it equal to zero and solve for x. Next, set up your intervals and plug a number on the intervals into your derivative (just like the first derivative test). *Note: If the problem says “non-negative” or something like that, start your intervals at 0. Anyway, once you plug in your numbers, decide what’s a max or min based on if you got a negative number or a positive number. Now, plug your x-values and your end points (such as 0 or, if stated, the sum) into your secondary equation and solve for the other variable. Then plug both variables into your primary equation. If you are looking for a maximum, your answer is the x-value with the biggest y-value. If you are looking for a minimum, your answer is the x-value with the smallest y-value.
Ex. Build a rectangular pen with three parallel partitions using 500 feet of fencing. What dimensions will maximize the total area of the pen? *x = width & y = length Primary equation: A = xy Secondary equation: 500 = 5x + 2y Solved secondary equation: y = 250 – (5/2)x Final equation: A = 250x – (5/2)x² Derivative: A’ = 250 – 5x Solved: x = 50 Intervals: (0, 50) u (50, 100) f’(1) = +ve f’(51) = -ve Plug in: If x = 0, then y = 250, and A = 0 If x = 50, then y = 125, and A = 6250 If x = 100, then y = 0, and A = 0 Result: y = 125 is the biggest Final answer: Your maximum area is 6250 ft²
In optimization problems we are looking for the largest value or the smallest value that a function can take. Here we will be looking for the largest or smallest value of a function subject to some kind of constraint. The constraint will be some condition (that can usually be described by some equation) that must absolutely, positively be true no matter what our solution is. The subject of optimization is kind of hard because a subtle change of wording can completely change the problem. There is also the problem of identifying the quantity that we’ll be optimizing and the quantity that is the constraint and writing down equations for each. The first step in any kind of optimization problem should be to read the problem carefully and identify the quantity to be optimized and the constraint. When identifying the constraint remember that the constraint is something that must be true regardless of the solution.
Ex: We need to enclose a field with a fence. We have 500 feet of fencing material and the side of a building is on one side of the field and will not need any fencing. Determine the dimensions of the field that will enclose the largest area. SOLUTION: 1. you will have two functions. The first is the function that you are trying to optimize and the second will be the constraint. 2. in this problem, we want to maximize the area of the field and we know that we will use 500 ft of fencing material. So, the area will be the function we are trying to optimize and the amount of fencing is the constraint. Maximization: A=xy Constraint: 500=x+2y We’re going to solve the constraint equation for x. X=500-2y We then substitute this into the area function leaving a function of y. A(y)= (500-2y)y=500y-2y^2 3. we want to find the largest value this will have on the interval [0,250]. Note that the interval corresponds to taking y=0 (no sides to the fence) and y=250 (only 2 sides and no width) 4. we’re going to use the critical points in the maximization formula. A’(y)=500-4 Setting this equal to zero and solving gives one critical point of y=125. Plugging this into the area gives an area of 31250 ft^2. 5. the problem originally asks for the dimensions of the field using the value of x the dimensions of the field is 250 x 125. We can prove this by plugging into the constraints formula X= 500-2(125)=250.
Works Cited http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx
Optimization is basically an extended application of what we have learned thus far in chapter 3 and some of chapter 5 if im not mistaken..
THE STEPS TO OPTIMIZATION: (are also listed on page 219 of our text book)
#1 list what is given to you in the problem
#2
list the primary equation
#3 List the secondary equation
#4
Solve for y in the secondary equation
#5
plug y= into the primary equation for y
#6
set = 0 and solve
#7
perform the first derivative test
meaning: take the derivative set = 0 solve for x set up intervals plug in for F(x) between intervals into the derivative determine if result is positive or negative and thereby increasing or decreasing respectivley on the intervals therefore enabling you to determine whther x= is a max or a min
EXAMPLE: A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. what dimensions will produce a maximum volume?
optimization problem are looking for the largest value or the smallest value that a function can take. Here we will be looking for the largest or smallest value of a function subject to some kind of constraint. The constraint will be some condition (that can usually be described by some equation) that must absolutely, positively be true no matter what our solution is. The subject of optimization is kind of hard because a subtle change of wording can completely change the problem. There is also the problem of identifying the quantity that we’ll be optimizing and the quantity that is the constraint and writing down equations for each. The first step in any kind of optimization problem should be to read the problem carefully and identify the quantity to be optimized and the constraint. When identifying the constraint remember that the constraint is something that must be true regardless of the solution.
hmmm i really dont know what optimization is. I went on wikipedia to try to get a lil info on it and it gave me a headache. but i know the suffix opti- means to see and ization- is the process of something. so im guessing optimization means the process of seeing something in calculus...
Optimization:
ReplyDeleteFollow the steps- The first thing you’re going to do is find the “given”, or the information the problem provides for you. Then you’re going to find a primary equation using the information they gave you. Next you’re going to find a secondary equation which is basically the given equation solved for either x or y. Then you are going to plug back into the primary equation. Then you’re going to take the derivative of the answer you just got from plugging in, and solve it for x. After doing that, you set up intervals to find if the function is increasing or decreasing. **Hint: if you make y 0 in the secondary equation it will tell you what the endpoint of your interval should be, which makes it easier to pick a point to plug into the derivative. If the interval isn’t going from –infinity, infinity then you’re going to have to check intervals by plugging all interval points into the given equation, to find the highest point. The highest point will be the answers.
Example: Find 2 non-negative numbers whose sum is 9 and so that the product of 1 number and the square of the other is a maximum.
1. Given: x+y=9
2. Primary: P=xy^2
3. Secondary: x+y=9
Solved: y=9-x
*Making y 0, x=9
4. Plug in: P=x(9-x)^2;=x^3-18x^2+81x
5. Maximize: 3x^2-36x+81; x=9,x=3
(0,3)u(3,9);f’(1)=+,f’(4)=-
Check:
X=0, y=9 P=(0)(9)=0;x=3,y=6 P=(3)(6)=18; x=9,y=0 P=(9)(0)
X=3, y=6 yields the highest product
**How do you do optimization?
ReplyDelete-Optimization basically means "maximizing" or "minimizing" so that means you're going to use the First Derivative Test. But before you can do that, you have to set up your own equations based off of the word problem. So the first thing you do is write down the formulas they give you and then determine which one is your primary formula and which one is the secondary formula. (*primary equation is the one that's closest to the words "maximum"/"minimum") Then you take your secondary equation and solve it for either x or y...I always solve for y, just to keep things simple. Then you plug your y into your primary equation and simplify it using algebra. Once you do that you take the derivative of the equation and solve for your x-values by setting it equal to zero. Then you set up your intervals, and lastly, you check your endpoints to see which one has the max/min value. Here's an example:
Ex) Determine the point(s) on y=x^2+1 that are the closest to (0,2).
*First, let's pull apart the directions and try to decipher what they mean
*When they say "closest" that means you're going to use the distance formula and it means "minimizing"..so you're looking for a min and you're going to use the first derivative test
*But first we need to plug into the distance formula. So asuming that the point you were given in the directions is x1 and y1, y2 must be the equation you were given in the directions. So when you plug in you should get this:
(x-0)^2+((x^2+1)-2)^2 ..*under a squareroot of course
*Now you're going to simplify that^ so you should get this under the square root:
x^4-x^2+1
*Now you can change it to (x^4-x^2+1)^1/2 because you have to take the derivative of it
*So deriving it you get (4x^3-2x)/2sqrt(x^4-x^2+1)
*Now take the numerator and set it equal to zero to find your x-values.
4x^3-2x=0
2x(2x^2-1)=0
x=0 x=+/-sqrt2/2
*Now you set up your intervals like this:
(-inf,-sqrt2/2)u(-sqrt2/2,0)u(0,sqrt2/2)u(sqrt2/2,inf)
*For the 1st interval you can plug in -1 and you should get negative, for the second you can plug in -.5 and you should get positve, for the third you can plug in .5 and you should get positive again, then for the last interval you can plug in 1 and you should get negative
*So that means at -sqrt2/2 you have a min, nothing at 0, and a max at sqrt2/2
*So since -sqrt2/2 is a min, you can plug that into the original equation to get the y-value for your point
*And you should get 3/2, so your final answer is:
(-sqrt2/2,3/2)
Source for^^^^^^^^
ReplyDeletehttp://tutorial.math.lamar.edu/Classes/CalcI/MoreOptimization.aspx
Okay, optimization is when you are maximizing or minimizing something. The first thing you need to do is read the problem and figure out if you’re minimizing or maximizing. Then, you need to figure out your two equations: primary and secondary equation. Now, your primary equation is the equation used for max/minimizing. And your secondary equation is the equation given to you in the problem. After this, you need to solve your secondary equation for one of the variables. Then, plug your solved secondary equation into your primary equation and simplify. Finally, we begin max/minimizing. First of all, you take the derivative of the equation you just simplified. Then set it equal to zero and solve for x. Next, set up your intervals and plug a number on the intervals into your derivative (just like the first derivative test). *Note: If the problem says “non-negative” or something like that, start your intervals at 0. Anyway, once you plug in your numbers, decide what’s a max or min based on if you got a negative number or a positive number. Now, plug your x-values and your end points (such as 0 or, if stated, the sum) into your secondary equation and solve for the other variable. Then plug both variables into your primary equation. If you are looking for a maximum, your answer is the x-value with the biggest y-value. If you are looking for a minimum, your answer is the x-value with the smallest y-value.
ReplyDeleteEx. Build a rectangular pen with three parallel partitions using 500 feet of fencing. What dimensions will maximize the total area of the pen?
*x = width & y = length
Primary equation: A = xy
Secondary equation: 500 = 5x + 2y
Solved secondary equation: y = 250 – (5/2)x
Final equation: A = 250x – (5/2)x²
Derivative: A’ = 250 – 5x
Solved: x = 50
Intervals: (0, 50) u (50, 100)
f’(1) = +ve
f’(51) = -ve
Plug in: If x = 0, then y = 250, and A = 0
If x = 50, then y = 125, and A = 6250
If x = 100, then y = 0, and A = 0
Result: y = 125 is the biggest
Final answer: Your maximum area is 6250 ft²
Source: http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/maxmindirectory/MaxMin.html
In optimization problems we are looking for the largest value or the smallest value that a function can take. Here we will be looking for the largest or smallest value of a function subject to some kind of constraint. The constraint will be some condition (that can usually be described by some equation) that must absolutely, positively be true no matter what our solution is. The subject of optimization is kind of hard because a subtle change of wording can completely change the problem. There is also the problem of identifying the quantity that we’ll be optimizing and the quantity that is the constraint and writing down equations for each.
ReplyDeleteThe first step in any kind of optimization problem should be to read the problem carefully and identify the quantity to be optimized and the constraint.
When identifying the constraint remember that the constraint is something that must be true regardless of the solution.
Ex: We need to enclose a field with a fence. We have 500 feet of fencing material and the side of a building is on one side of the field and will not need any fencing. Determine the dimensions of the field that will enclose the largest area.
SOLUTION:
1. you will have two functions. The first is the function that you are trying to optimize and the second will be the constraint.
2. in this problem, we want to maximize the area of the field and we know that we will use 500 ft of fencing material. So, the area will be the function we are trying to optimize and the amount of fencing is the constraint.
Maximization: A=xy
Constraint: 500=x+2y
We’re going to solve the constraint equation for x.
X=500-2y
We then substitute this into the area function leaving a function of y.
A(y)= (500-2y)y=500y-2y^2
3. we want to find the largest value this will have on the interval [0,250]. Note that the interval corresponds to taking y=0 (no sides to the fence) and y=250 (only 2 sides and no width)
4. we’re going to use the critical points in the maximization formula.
A’(y)=500-4
Setting this equal to zero and solving gives one critical point of y=125. Plugging this into the area gives an area of 31250 ft^2.
5. the problem originally asks for the dimensions of the field using the value of x
the dimensions of the field is 250 x 125.
We can prove this by plugging into the constraints formula
X= 500-2(125)=250.
Works Cited
http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx
Optimization is basically an extended application of what we have learned thus far in chapter 3 and some of chapter 5 if im not mistaken..
ReplyDeleteTHE STEPS TO OPTIMIZATION: (are also listed on page 219 of our text book)
#1
list what is given to you in the problem
#2
list the primary equation
#3
List the secondary equation
#4
Solve for y in the secondary equation
#5
plug y= into the primary equation for y
#6
set = 0 and solve
#7
perform the first derivative test
meaning:
take the derivative
set = 0
solve for x
set up intervals
plug in for F(x) between intervals into the derivative
determine if result is positive or negative and thereby increasing or decreasing respectivley on the intervals
therefore enabling you to determine whther x= is a max or a min
EXAMPLE:
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. what dimensions will produce a maximum volume?
#1
given: SA=108in^2 SA= x^2+4xy
Primary Equation: V= x^2y
Secondary equation: x^2+4xy=108
4xy=108-x^2
y= 108-x^2/ 4x
Plug in: V= X^2(108-x^2/4x)
V= 108x-x^3/4
27x-1/4x^3
first derivative test:
27-3/4x^2=0
-3/4x^2= -27
x^2=36
x= + - 6
Set up intervals
(0,6) U (6,sqrt108)
F(1) is positive and thereby increasing
F(7) is negative and thereby decreasing
Therefore X=6 is a maximum
Therefore V= LXWXH
V= 6X6XH
6^2 x y = 108
y= 108/ 36
y=3
V= 6X6X3
optimization problem are looking for the largest value or the smallest value that a function can take. Here we will be looking for the largest or smallest value of a function subject to some kind of constraint. The constraint will be some condition (that can usually be described by some equation) that must absolutely, positively be true no matter what our solution is. The subject of optimization is kind of hard because a subtle change of wording can completely change the problem. There is also the problem of identifying the quantity that we’ll be optimizing and the quantity that is the constraint and writing down equations for each.
ReplyDeleteThe first step in any kind of optimization problem should be to read the problem carefully and identify the quantity to be optimized and the constraint.
When identifying the constraint remember that the constraint is something that must be true regardless of the solution.
optimization????
ReplyDeletethat sounds like something that I've never heard of........so I'm just gonna end it off here............
hmmm i really dont know what optimization is. I went on wikipedia to try to get a lil info on it and it gave me a headache. but i know the suffix opti- means to see and ization- is the process of something. so im guessing optimization means the process of seeing something in calculus...
ReplyDelete