Thanks man, you are a really amazing at teaching this stuff. I am a first year uni student and this really helped me a lot (maths at uni is not as hard as many people think). Keep up the great work :)
Thanks so much for all of the work you put into these videos! ^_^ One random question for you though...how many Sharpies do you think you've gone through making all of these? XD
I did this problem differently than you and arrived at a different answer. I'm inclined to believe you over me, but I don't understand why my method doesn't seem to work, assuming I'm not making a careless error in taking my derivative. Basically, if you have a 20 in wire, you cut it in half and designate some piece as x and the other piece as 20 - x. Each piece is then used to form a square. One square will have side lengths of x/4 and the other square will have side lengths of (20 - x)/4. Work it all out, and it seems that x = 10, and the end result is that the wire is cut in half. You're much better at math than I am, but I cannot figure out why this method isn't working.
Since there is only one derivative = 0 and it is a minimum. The maximum must be at the ends of the function. X=0 and X=20, in the case. In other-words putting all your wire in one square and making the other zero in =size would be maximize the area.
you have to do the first derivative test, when you put the derivative to zero, that gives you POSSIBLE maximum OR minimum values, to see which one it is (max or min.) you have to do the first derivative test and check the intervals AND he has a video on that called first derivative test and critical numbers lol it will rlly help if you're confused with how to know the difference
In the other examples, taking the derivative of setting it to 0 was the maximum.. How come this is the minimum in this case? How can you know the difference?
BennduR They both have the same steps there is no way to know until you find when the derivative=0 after you find that point. You take one point before and after and sub it in the derivative and see if it's inc. Or Dec. After the point and that means that function is either an absolute max or min at the point when the derivative=0.
The maximum is also equal to the minimum in this case. If you imagine one large square and one infinitesimally small square (using no material) then the area would also be 25 ft^2.
I've whatched all of your optimization videos, thank you so much, i definitely learned the logic.
I just learned a concept in 7minutes, that my teacher lectured on for 1 hour
Thank you for having this channel you lifesaver!
You literally saved my life!! Thank you so much for all these videos CLEAR and different each time with different levels 🙌
Émilie Boucher by the way did that hurt?
Thanks man, you are a really amazing at teaching this stuff. I am a first year uni student and this really helped me a lot (maths at uni is not as hard as many people think). Keep up the great work :)
Doing math with a sharpie... Bold move.
Patrick never makes mistakes
Thank you so much for making these optimization problem videos! They are extremely helpful! :)
This is probably one of the easier optimization problems. I seriously hope my test doesn't have something like the 5th and 6th example. UGH
Or the infamous 3rd!
OMG THANK YOU SO MUCH. KEEP UP THE GOOD WORK. JUST SAVED ME
Thanks so much for all of the work you put into these videos! ^_^ One random question for you though...how many Sharpies do you think you've gone through making all of these? XD
I did this problem differently than you and arrived at a different answer. I'm inclined to believe you over me, but I don't understand why my method doesn't seem to work, assuming I'm not making a careless error in taking my derivative.
Basically, if you have a 20 in wire, you cut it in half and designate some piece as x and the other piece as 20 - x. Each piece is then used to form a square. One square will have side lengths of x/4 and the other square will have side lengths of (20 - x)/4.
Work it all out, and it seems that x = 10, and the end result is that the wire is cut in half. You're much better at math than I am, but I cannot figure out why this method isn't working.
my exam review right now. thanks a lot!:)
omg im so glad to have found your channel! thank you so much!!
Since there is only one derivative = 0 and it is a minimum. The maximum must be at the ends of the function. X=0 and X=20, in the case. In other-words putting all your wire in one square and making the other zero in =size would be maximize the area.
you have to do the first derivative test, when you put the derivative to zero, that gives you POSSIBLE maximum OR minimum values, to see which one it is (max or min.) you have to do the first derivative test and check the intervals AND he has a video on that called first derivative test and critical numbers lol it will rlly help if you're confused with how to know the difference
thank you so much for these videos! very helpful!!!
@theprocrastinator100 yep! turns out it would be
In the other examples, taking the derivative of setting it to 0 was the maximum.. How come this is the minimum in this case? How can you know the difference?
BennduR They both have the same steps there is no way to know until you find when the derivative=0 after you find that point. You take one point before and after and sub it in the derivative and see if it's inc. Or Dec. After the point and that means that function is either an absolute max or min at the point when the derivative=0.
You take the second derivative and use the "second derivative test."
Definitely OVER 9,000!
@Twinsfan36 thanks! : )
How could you find the maximum area?
The maximum is also equal to the minimum in this case. If you imagine one large square and one infinitesimally small square (using no material) then the area would also be 25 ft^2.
Is there anyway to figure out the area of the two boxes, or would you need more information?
whether it is to maximize or minimize, i have to equate the derivative to 0? im confused :/
When doing the first derivative do you need to use chain rule or can you expand then derive?
could you maybe do one like this but finding min and max. and using a square and circle?
I think it helped but can you do one for just one square because that one is not coming out as easy as the one you did
CLASSIC!!
How do u know when to use different variables?. I used x for both squares and ended with a diff equation! thx
When the derivative is zero there is a flat spot on the original function which creates maxima and minima
maximize the volume of a rectangular prism with surface 100
How would a person maximize this problem?
Have the first square made have an area of almost 25 and the other square have an infinitely small area
What if the question said find the maximum?
well, at a glance this should do that as well. the max should occur at an endpoint in this example which would give the same solution (try it).
@@patrickjmt i get x =5 and y =0 so the area would be 25. But the problem is the wire is not cut.