I ust cancelled the x^2 on the bottom by multiplying it on both sides to get to 2x^3 - 128000 = 0, because x can't be zero in this case (not a solution), and then dividing by 2 and noticing that 64000 = 64 * 1000 which is the product of two cubes, namely 4 and 10, thus the answer is 40
But if he had added 128000/x² tot he other side, he would have gotten the right answer within 3 seconds. His method took over 10 seconds. In the ap test, time is literally money, since you paid to potentially save hundreds on another college class
Hi! I just found a constant (approximately 2,2911249) by adding up "infinitely times" iterated heights beginning in a 60-60-60 degree triangle, then iterating it in 30-60-90 degree triangles so that the height lines were perpendicular to the hypotenuses (and in the end I found a point that these lines were approaching). The question is the following: does this number already exist, and if yes, then how is it called? Sorry, if my explanation was kind of messed up, but I'm not that good at english. Thanks for the answers in advance. :)
Let x^2*y = a^3 be fixed. We consider the minimum of x^2 + 4xy. We note that x^2 + 4xy = x^2 + 4x( a^3/x^2 ), = x^2 + 4a^3/x, = x^2 + 2a^3/x + 2a^3/x. By AM-GM inequality we have ( x^2*2a^3/x*2a^3/x )^( 1/3 )
Well, since there are two dimensions which we can "act" on (x and y), the endpoints would be : - when x is very big and y is very little (so y = 0, but y = 32000/x² can't be 0 for a finite value of x) ; - when y is very big, so when x is very little, but x ≠ 0 because y = 32000/x² would be undefined Hence there isn't any endpoint here
I think the closest you could get to actually checking the endpoints here would be to find the limit of A(x) as x approaches 0 and infinity. As noted by Monmamat, x can't actually take either of these values as both would cause the volume to no longer be fixed, but the limits will give us an idea of what happens to the area as we get as close as possible. But you'll find that the limit in both cases results in the box having infinity surface area, and so they're clearly not candidates for the minimum.
I ust cancelled the x^2 on the bottom by multiplying it on both sides to get to 2x^3 - 128000 = 0, because x can't be zero in this case (not a solution), and then dividing by 2 and noticing that 64000 = 64 * 1000 which is the product of two cubes, namely 4 and 10, thus the answer is 40
God, I remember having this question on a Final for Calc1 Applications. Still gives me nightmares!
at 4:40 why the hell would you solve it that way
i was shocked as well, only thing i can think of is that it helps when checking values to see if there is truly a min/max lol
But if he had added 128000/x² tot he other side, he would have gotten the right answer within 3 seconds. His method took over 10 seconds. In the ap test, time is literally money, since you paid to potentially save hundreds on another college class
Now do something with multivariable and use partial derivative
Hi! I just found a constant (approximately 2,2911249) by adding up "infinitely times" iterated heights beginning in a 60-60-60 degree triangle, then iterating it in 30-60-90 degree triangles so that the height lines were perpendicular to the hypotenuses (and in the end I found a point that these lines were approaching). The question is the following: does this number already exist, and if yes, then how is it called? Sorry, if my explanation was kind of messed up, but I'm not that good at english. Thanks for the answers in advance. :)
Let x^2*y = a^3 be fixed. We consider the minimum of x^2 + 4xy. We note that
x^2 + 4xy = x^2 + 4x( a^3/x^2 ),
= x^2 + 4a^3/x,
= x^2 + 2a^3/x + 2a^3/x.
By AM-GM inequality we have
( x^2*2a^3/x*2a^3/x )^( 1/3 )
You look so good with hair!
since we usually check the endpoints in a candidate test, what would be the equivalent of the endpoints in this question?
Well, since there are two dimensions which we can "act" on (x and y), the endpoints would be :
- when x is very big and y is very little (so y = 0, but y = 32000/x² can't be 0 for a finite value of x) ;
- when y is very big, so when x is very little, but x ≠ 0 because y = 32000/x² would be undefined
Hence there isn't any endpoint here
I think the closest you could get to actually checking the endpoints here would be to find the limit of A(x) as x approaches 0 and infinity. As noted by Monmamat, x can't actually take either of these values as both would cause the volume to no longer be fixed, but the limits will give us an idea of what happens to the area as we get as close as possible. But you'll find that the limit in both cases results in the box having infinity surface area, and so they're clearly not candidates for the minimum.
Are these from The Archives or are you aging backwards nowadays
meow