Tangent line toa curve that minimizes the area of a triangle

[rocomath]

Lol, I made that mistake too but I realized no that can't be right since I have to move 2x to the other side :-]

But yeah, that was a good problem.

 

[NyteBlayde]

REPOST: wasnt sure if ud see it no since this is page 3 lol
thanks a ton for helping out on that one :D

back on the topic of the first question, u set 1/2(1-3x^2) = 0

so i should end up with 1/2 - 3/2x^2 = 0 and all i need to do is solve for x and then find my min right?

 

[rocomath]

Instead of multiplying the half through, just divide and get rid of it. Then yes just find where x=0, which is a critical point and plug it back into your original.

 

[NyteBlayde]

lol this going to be kinda ugly, +/- 1/.sqrt 3 (without rationalizing)

 

[rocomath]

lol this going to be kinda ugly, +/1 1/.sqrt 3 (without rationalizing)

Yeah that's what I got too, but remember, x=base which is representation of length so we take only the positive value.

 

[NyteBlayde]

yuck i hate dealing with radicals lol

ok, i think i got it. if I am right, the extremes of the area are at P(.sqrt3 / 3 , .sqrt3 / 9) and Q( -.sqrt3 / 3 , -2.sqrt3 / 9 )

 

[NyteBlayde]

Alright, once again thanks for all your help lol, couldn't have done it without you :D. Now i can finalize these in word and put them away :)

 

[rocomath]

You're only value should be $$x=\frac {\sqrt3}{3}$$ and plug in it $$y=1-x^2$$ for your corresponding y-value which is your height.

 

[NyteBlayde]

lmao i feel so retarded, i pluged it into the area function instead of the original >.< good thing you caught me

thx again :)

 

[rocomath]

yuck i hate dealing with radicals lol

ok, i think i got it. if I am right, the extremes of the area are at P(.sqrt3 / 3 , .sqrt3 / 9) and Q( -.sqrt3 / 3 , -2.sqrt3 / 9 )

lmao i feel so retarded, i pluged it into the area function instead of the original >.< good thing you caught me

thx again :)

I'm not sure if you read post #35, but for this problem you can only take the positive x values b/c x represents the base which is a measure of the length of the triangle and it can only be positive.

Also, you're corresponding y-value should also be positive b/c you want to minimize this triangle so that it is in the 1st quadrant.

If all your values are positive, that means you probably did the question correctly!

It satisfies the physical conditions along with the xy coordinate system.

 

[NyteBlayde]

I know, i just did both extrema, i only needed min and there was an extraneous value, just wanted to do the math

 

[rocomath]

I know, i just did both extrema, i only needed min and there was an extraneous value, just wanted to do the math

Lol, damn you're hardcore.

 

[NyteBlayde]

Lol, damn you're hardcore.

damn straight :D

 

[rocomath]

AHHH! Lol, round 2 to fix the error?

$$y'=\frac{-(2x+y)}{x+2y}$$

$$y'=0=-(2x+y)$$

$$y=-2x$$

OHHH! Big break :-] I guess I can't trust my intuition.

 

[NyteBlayde]

lol round 2 was short lived :P same damn answer lmao

 

[rocomath]

lol round 2 was short lived :P same damn answer lmao

Yeah lucked out, lol. It worked out sweet since both terms were the same sign, whew! I was like damn all that work for nothing, and plus I hate to leave things incorrect.

 

[NyteBlayde]

agreed, that's why i figured when i found the error while writing it out i figured id tell you :)