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Prinay

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Problem #2

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Attempt For # 3

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This was my attempt for #5. I wasn't sure how to integrate the e. Did anyone do anything different?

do we have to apply the antiderivative to the e too?

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Question #3

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The steps are easy to understand and do but I just need to go over it a few times to get used to it.

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The critical points are when the derivative is undefined or 0. This will help find the minimum and maximum points.

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The points show maximum and minimum points of the function

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A similarity would be that they have the same step process and the first and second derivatives are solving for 0

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1. Making an arbitrary guess of width and height of of 600 ft. of fencing.
2. Identify constraints and an area and perimeter formula to later interchange values through substitution.
A(x) = W x H
P = 2H + 3W = 600 ft.
W greater than(>) or equal(=) to 0
H greater than(>) or equal(=) 0
3. Using perimeter formula solved  for the height in terms of the function.
H = 200 - (2/3)W
4. Plugged new found value H into the area formula identified earlier.
A(W) = W x ( 200 - (2/3)W)
= 200W - (2/3)W^2
W less than (<) 300ft.
5. Calculated derivative of current area function to maximize the width.
a. Solved for Critical Points
A'(W) = 200 - (4/3)W
= 150
b. Determined critical point was a local maxima
A'(100) = 200 - (4/3)(100)
= 66.66
A'(200) = 200 - (4/3)(200)
= -66.66
6. Solved original function for 150 to find the area and divided the area by the height to solve for the width.
7. The pen with the largest area would have a width of 100 feet, a length of 150 feet, and an area of 15,000 square feet.

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1. Draw a picture of the scenario
2. Create an equation that relate the variables and constraints.
3. Use equality constraints to solve for one variable.
4.  Find the critical points (Where the derivative is 0)
5. Use the first or second derivative test find the local min and max
5. Find the global extremum.

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1. model the scenario
2. create an equation that represents the situation (include constraints + variables).
3.  use equality constraints to solve for one variable
4. find the critical points (where the derivative = 0)
5. use the first OR second derivative test to determine which extrema are local min/max (if they exist)
6. find global extremum

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Prinay
Notice: Using the quadratic formula for finding the critical points
Wonder: How will the second derivative test will play in this problem?

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I notice that I got the same answer by inputting w = 14 into the equation and then checking my work by graphing.

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I noticed that the quadratic formula was used

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I noticed the usage of graphing functions. using the quadratic formula and applying that to formula provided in the assignment. I also noticed I got the same answer. I wonder what purpose the of the quadratic formula serves in this process of solving the question and why you decided to use it?
-Felipe Delgado

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I noticed Cho used the derivative and found the critical points.

I wonder how I can use your example to revise my summative.

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Notice : the derivative and critical points are used for this version
Wonder : is this method better or more correct than the other method shown?

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I notice that the critical points were found of the function.

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Notices: Finding critical points
Wonders:
Didn't know that we could say that the points are from Desmos.
When using the quadratic equation I couldn't find the critical points, was there a way to

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I notice that Cho found the critical points of the function to then see whether the maxima or minimum wage was the better choice.
Something I wondered before looking at this example was what were the endpoints, and in this case 14 was the endpoint.

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I noticed that the first steps taken were to find the critical points of the function and then use the first derivative test

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Notice:
-  Used the first derivative to find critical points.
- Used the quadratic formula
- This has more steps but it leads us the same answer I have.

Wonder:
- How can we use optimization to figure out a problem like this?

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notice:
Although there were more steps involved, generally, the answer was the same as the one I got

wonder:
can there be more than one way to use optimization to figure out an answer?

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I notice that Cho used desmos to graph the function & check his answers.

This looks more like the calculus we've been practicing.

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I notice that the quadratic formula was being used. Critical points were used to find the max and min.

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Notice
- found critical  points to figure max which explores the possibility of a greater optimum wage that can employ more floor workers
- used quadrative formula b/c polynomial couldn't be factored
- used Desmos to cross reference

Wonder
- Why did you choose first derivative test over second derivative test when it's quicker?

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Notice: Used 1st derivative and critical points
Wonder: Why was the quadratic formula used?

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Notice:
- In the graph there is a function
- The derivative function is graphed
- I noticed the max/mins matched up with zeros
- x=100 was the point of inflection, it looks like
- Cho took the derivative and used the quadratic formula to find the zeros of the function
- He also did it another way; he used the line test to figure it out
Wonder:
- Did cho visually indicate in the graph where y = 14 (the minimum wage)

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Notice: Used the derivative to find the critical points and used the quadratic formula.
Wonder: why did you use the quadratic formula?

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The first step for optimization produce is to draw visuals to represent the problem. The second step is to create an equation which is A(x) = x1 x2. The third step is to use the constraints to help you solve the equation. The fourth step is to find the critical points of the function. The fifth step is to use either the first derivative test or the second derivative test to help you find the max and min values of the function. The last step is to find the endpoints of the function to get your final answer of the optimization process.

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1) Label and sketch your problem with the information given to you
2)formulate the A(x)- variable equation and solve for it.
3)Find critical points by using first and second derivative test, use sign  chart (determine local min or local max).
4)plug in local max/min, to find final answer.

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Traveling Optimization can be used from finding one way to another and determining which path is the shortest to take.
Shopping can be used to maximize quality of the product is made from and minimizing the pricing.

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Step 2 and 3 are confusing steps to me. Setting up the equation is kinda hard and confusing to do but I guarantee if I work on it I"ll be able to get it.

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1) We first draw a picture of the problem and label them.
2) We then create a formula/equation A(x)- variable and constraints and solve for it.
3)We look for the critical points by using derivatives and with a sign chart, we were able to identify if the point was local max or local min.
4) We plug in the local max/min in our area function to find the local extrema.

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First, draw a model for the scenario. Next, come up with an equation (in this case, for the area, so it would be A(x) = length x width). Write the constraints, to rule out the numbers that the length and width can't be, and to also include what values you already know (like the perimeter). Then, use the equality constraint to solve for one of the variables, and then using substitution, get rid of that variable so you're only left with one other variable in the equation, making it possible to solve. And then once you have the original equation (the A(x) one), but with only one variable instead of 2, take the derivative, find the critical point, and use a sign chart to figure out if that point is a local max or local min. Then, take that critical point and plug it into the area function to find out the global extrema. And then also check the end points (in this case, it's the points where the area would equal 0), to make sure those are the global max or min.

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1) Draw a picture that's labeled with the information you are given
2) Create an equation that includes the variables and the constraints
3) Solve the equation so that you're left with less independent variables
4) Find the critical points by using the first derivative test or the second, if it calls for it
5) Plug in the critical points into the first or second derivative to figure out if the points are max or mins
6) Then find the global extremum by plugging in the critical points and/or endpoints

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First you draw a picture of the problem and label it. Then you create an equation with constraints and solve it. Next you find the critical points by using the 1st and 2nd derivatives. Finally you use those points to find either the maxima or minima.

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First, draw a model of the problem. Then, we have to formulate a equation A(x)- variable and constraints. Then, we solve the equation that we created. After than, find the critical points using the first derivative and second derivative test and find the sign chart to find out if it local maxima or minima. Finally, then plug in the maximum or minima in the equation to get the answer.

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First, we draw the scenario for visualization. Then we create equations like the objective function, that relate to the variables and constraints. We then use the equality constraint to reduce and solve for  one independent variable, and substitute it. Next we find the critical point of the objective function and test if they are local minimums or maximums. Finally we compare the objective function values at local extrema and at the endpoints  to find the global extremum or "goal".

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2.  Formulate an equation with constraints
3. Solve the equation working with the independent variable.
4. Find the critical points by using the 1st and 2nd derivative test.
5. Using the critical points find the minimum and maximum
6. Then find the global extremum
7. Use a graph to check

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1. Draw a picture and label with information you know.
2. Create equations with variables and consider constraints.
3. Reduce the number of variables and substitute into objective function and inequality constraints
4. Use power rule to find critical point
5. Use first or second derivative test to find local max and min.
6. Plug in critical point into objective function

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Begin by drawing a model for reference. Then, create the objective function ( the A(x) one). Determine if you are trying to maximize or minimize. Determine the inequality constraints and then the equality constraint (perimeter). Solve for one independent variable (x or y). Use substitution in order to remove the independent variable from the objective function and inequality constraints. Find the critical points by using the first or second derivative test of the objective function. Determine if the critical point is a local maximum or minimum. Compare the endpoints in order to find the global extremum using the area function. Determine the maximum amount of space and the length and width. Lastly, confirm it using a graph of the area function.
- Felipe Delgado

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First to optimize a function, what you have to to formulate the equation, as well as come up with constraints, as well as another function that represents the values you are working with. Then you want to work with the last iv, trying to find its value, and then plugging it back to the original function. After that, what you want to do is set the equation up as an inequality (x2), to find the last value, where there is zero. Then you want to find the first derivative and the second derivative, to find the critical points, the maxima and minima, and well as its concavity to find the maximum value. Then you want to plugin in the critical point to the original improved function to find the maximum value.

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Similarities: uses number line to determine if it concave up or down, (uses power rule) equals to zero.

Differences: first derivatives are for critical points, second derivatives are for concave.

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First derivative is the test to find out the maximum and minimum points. which are maxima and minima.

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Red: concave up
Blue: concave down
purple: concave up
orange: concave down

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Blockbuster was an inflection, due to the fact that they are actually out of business and there are apps or places that you can stream movies from.

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The equality constraints are somewhat confusing, but everything else makes sense.

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Prinay
What I understand are drawing a model to represent optimization and solving for one IV. I don’t understand how to use substitution to remove IV from the function and the constraints that are inequalities.

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The second part is a bit confusing but after re watching the video I am sure that could be fixed

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The math is clear. But similar to what Elyna said why can’t we use the original area formula.

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I'm just confused on the constraints . Like I understand what we have to do but how will we know when to use it? (if that makes sense??)

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I'm kind of confused, like all the steps make sense and everything, but in general it's just kind of complex and I think if I look at more examples I'll understand it better.

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I am confused on the formula why we use that one.

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I think steps 2 and 3 can get a little confusing for me, more specifically the part were we solve for one constraint and then use substitution

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I guess I would be confused for the maximizing and minimizing for the inequality.

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its a bit confusing that we were talking about the area and plugging in the values for the area, and then started talking about the limitations again and doing more things with those.

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I'm kind of just confused on it all in general. Well mostly kind of around steps 2 and 3, but kind of still confused altogether

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I understand what we have to do, but I'm a bit confused on step 3 (constraint).

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Why won't the regular area formula work? The math is understandable the why is confusing.

Do we have to remember geometry to do this?

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I am confused on the second step a little. I understand the third step on how to set the equation and the substitution.

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I understand how the optimization process works with 2D area and perimeter, however identifying inequality restrictions is definitely a growth area. The reason for each step is also unclear. I'm curious to learn about how it can be used with 3D figures.

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The steps we went over make sense after being clarified especially by doing close reading as it helps me to understand better.
-Felipe Delgado

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I understand the substitution and how the questions are formed. The only confusing thing is the difference between constraints that are equal and unequal.

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I am confused in the 3rd part of the problem where we have to reduce and substitute and then come up with inequality restrictions.

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I think what is confusing is maybe trying to see what the difference between maxing and minimizing will affect your equations and what you're solving. I understand the restraints based on the situation, but do we have to memorize any new formulas when forming the function with the independent variables.

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When shopping, maximize quality minimize price

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Finding the right bus routes to take in order tho spend the least amount of money and time.

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It can be used for transportation, finding a way to get from one point to another, where the path is the shortest path you can take.

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optimizing code for computers so that the program runs the fastest without using as much processing power

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Prinay
A real world example of optimization is when any other transportation services optimize the schedule of the time.

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When building you would use optimization to build efficiently - maximizing build time, but at a lower cost

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I think that in traveling optimization is used. If you're on the road you want to choose the fastest route to save time and gas. Then there's the concept of trying to do the most in a day.

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An example would be profit in a business

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when traveling, you would be trying to maximize the distance while also minimizing the cost

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When looking at different farming techniques. Are you focusing on maximizing the amount of produce, or the quality

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Traveling

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You can use optimization on schedules in order to be more efficient

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Optimization can be used to maximize profits and minimize production cost.

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Optimizing things like underground fiber and the area it covers and where it's going to go would need a lot of optimization to get the best results.

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When practicing yoga we can maximize our strength, flexibility and mobility, and reduce/minimize sore inflexible muscles.

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An example where we might use optimization would be traveling and using some sort of schedule.

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Determining routes to different aircrafts

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traveling by car

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When handling water distribution through pipes, you need to calculate the exact amount and rate water is being filtered. You might use optimization to make sure that the water is being filtered safely in the quickest rate possible

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Stretching : Maximize flexibility and minimize any tightness or risk of injury. -Felipe Delgado