[exitwound]

Originally Posted by masraum

OK, I was correct about the beginning, and I suspect that's where you're getting hung up. Normally, the area of a triangle is ½bh. Since they say "equilateral triangle" we know the b and h are equal so Area of an equilateral triangle is ½s²

There will be 4 variables, A, A', s and s'.

A = given (16in²)
s = working backwards from 16in² gives 4√2
A' = 4 in²/min
s' = don't know, that's the problem.

A = ½s²
A'= ss'

s' = A'/s = 4/4√2 = 1/√2 = 2/√2

The side is increasing at 2/√2 in/min

Actually the answer is 1(3^1/4) in/minute.

• Yes, the area of the equilateral triangle is ½s².
• s=4√2
• dA/dt = 4 in²/min
• So if A=½s² , dA/dt = 2(½)s(ds/dt)
• simplify: dA/dt=s(ds/dt), 4=4√2(ds/dt),
• ds/dt = 1/√2 which is WRONG and why I can't figure out how to do this problem and want to stab someone in the throat.

Hmm that Area equation is probably where I was going wrong too.... WHY WAS GEOMETRY 17 YEARS AGO?!