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I don't get it... can anybody help please?
you have to integrate on root(x) from x=0 to x=9.
actually that doesn't give you a rectangle. hmmm....
screen shot is nothing
hmm, i can't see anything
you have to find the largest rectangle possible. looking at the graph i would guess (4,2) but that's just a though....
apparently we have to do it through optimization - derivatives and all that fun stuff
This is one of those questions you ask an instructor whose got over 20 yrs cal experience.
find a function for the area of the square and optimize. the area of the square at a given x up until 9 = x*y = sqrt(x) * (9-x)
try deriving that and see if find an extrema
not really :) as long as you have the skill then it's no prob
it's one of those questions that are a challenge ^_^
loving the enthusiasm sstarica, but i'm getting mercilessly owned in cal a
LOL, nah, let it be the opposite. It's quite simple, believe me :)
logically, I'd say the following: since the lenght of the rectangle is = 9, and is under the sqrt(x) now, let's first draw sqrt of (x) and notice that when you take x = 9, you'llhave y = 3. Logically, the rectangle won't exceed this limit, so the width is equal = 3
matt has ir right! good job
since it's inscribed under it
yes the maxima of that graph is 3!
:) no need for all this.
now we need to find A'
all you had to do is draw sqrt(x) and see the limit of the rectangle. ^_^
why A'? doesn't he want A?
A = 3 x 9 = 27 that if he asked for A right? ^_^
to maximize you find A'
so it's 3 * sqrt(3)
or no. 6 * sqrt(3)
oh right lol my bad
yes x=3 is right matt thats are only critcal number thats in the domain of A
wait wait, why is 9=xy? doesn't A=xy?
since he wants the largest area, then yes maximize, proceed matt and hope you understood what was going on Ire :)
i didnt see A=9 up there no
you want the largest area, so you have to maximize in this case, they didn't ask for the "area" alone, but largest one
right. to find where the area is a maximum, take a function for the area and find the maxima point. take that point to make your rectangle. since you go up to 9 the x width is (9-3) = 6. the height of the rectangle is sqrt(3)
could you remind me how to find the maxima point please?
which means "optimize" lol, alight matt will take it from here ^_^ good luck
take the derivative of the area function 'A'. find where the derivative = 0.
normally, you'd have to do the second derivative test to find out if it is a max/min too. but i know secrets.
1. A = (9-x) * sqrt(x)
wow, thanks guys
also i never use the second derivative test it is pointless you can use A' to see if it is increasing to decreasing at x=3 to find that it is a max
The first derivative rocks!
thanks for the help guys, really appreciated