Optimization Calculus – Fence Problems, Cylinder, Volume of Box, Minimum Distance & Norman Window

Optimization Calculus – Fence Problems, Cylinder, Volume of Box, Minimum Distance & Norman Window

49 Replies to “Optimization Calculus – Fence Problems, Cylinder, Volume of Box, Minimum Distance & Norman Window”

  1. I FUCKING LOVE YOU. I have a college calculus exam tomorrow on Mean Value Theorem, Curve Sketching, and Optimization, and I was out with the flu for the entire two weeks that my professor taught it. I now know everything I need to do and you sir, have genuinely saved my life. If I fail this class, I will be dismissed from college. You prevented that from happening. I will repay this to you someday.

  2. this channel is the sole reason I've passed any math class in college….and I'm about to use this in ochem too lmao

  3. 20:41 Instead of making the pen into bars, you could turn it into a window shape. Giving 3x+3y=750, then both x and y turn out to be 125ft making the largest area possible 15,625 sq. ft. I've learned a lot from your videos btw, THAANKS!

  4. Question, with the largest area of the 4 pens, if you have 3x+3y as the constraint function (so that there is a cross divide in the middle) the total area is 15625ft^2, isnt this the correct answer? edit: 18:00

  5. 16:57 cant you just do p=2x+2y solve everything and once you get the area, divide the whole thing by 4?

  6. I got a different solution for the example with the 750ft. fence and 4 pens. it is possible that the pens are not next to each other, but two and two, forming a square.

  7. HI everyone, I have a question for the exemple at 17:00. He separate the rectangle with 4 lines, which give him this 2x+5y=750 for the perimeter. I was wondering if instead we can separate like a cross, which will give 3X+3Y=750. It is now more easy, you got 3(X+Y)=750. Divide both side by 3 and you got X+Y=250. Than solving Y=250-X, we got x*(250-x)=250x-x^2. first derivative is 250-2x. Solve for x. x=125.
    3*(125) +3*(125)=750
    A=125*125= 15625 ft^2
    Which is bigger than 14062.5ft^2

    did I make a mistake? Can you help me figure it out?
    English is my second language, sorry for my mistakes.

  8. For the first couple problems you are finding the local min/max but you also have to test endpoints of domain for absolute max/min

  9. for the problem with splitting the fencing into 4 pens could you have it with 3 vertical and 3 horizontal to form a window shape? I did this and got 15625 ft^2 but im not sure if you can have the overlap in the middle.

  10. Suggestion, instead of one long video wouldn't you like to make shorter videos tackling problems instead of daunting some people with very long ones when say we're stuck with a specific question but don't really want to watch the whole thing? Thanks

  11. How do you know if you're minimising or maximising a function if you use the identical method of setting the gradient to 0? For example in the cylinder example, how do you know you're minimising the dimensions and not maximising them?

    Also, how would you maximise the surface area if you wanted to? I assume that the equation formed would be identical too.

  12. isn't it that the square root of 400 gives a positive and negative 20 values of x? and when you solve for y it also gives positive and negative values of y. therefore when you identify the coordinates of the minimum it should be (-20, -20)

  13. It's funny that my calculus teacher just give us a a quiz assignment with this and he only copied it on YouTube only change the numbers and some sentence question#3

    Hi sir pherie if u read this
    I want to say us students watch YouTube 2😂😂😂😂

  14. For the fence problem, with 4 pens, why not (750/6)^2=15,625? Thank you for making videos to make me think in new ways.

  15. I had a question at 9:07

    When we finish the square root , can we pick x in order to be -20 so that the sum will be a minimum?

  16. This video was a lifesaver thank you. I could not figure out how to optimization problems for the life of me. I failed that question on the last calc test. I have my exam coming up soon and I wanted to make sure I understood it and now I do! Thank you

  17. First five problems can be solved easily by using "Cauchy's Inequality": (a+b)/2 >= sqrt(a.b) where a, b >0. Product a.b (or area) a.b is max, or Sum a+b (or perimeter) is minimum when a=b. For problem 1: a + b=60 so a.b is max when a=b=30; problem 2: a.b=400 so (a+b) is min when a=b=20; problem 3: perimeter is 100, so a+b =50, thus area is max when a=b=25; problem 4: x+2y=4000, so area is max when x=2y=2000 or x=2000 and y=1000; problem 5: 5y+2x=750, so area is max when 5y=2x=350, thus y= 75 and x=350/2.

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