Find the area of the largest rectangle with a base on the positive x axis, its right side on the line x=9 and which is inscribed under the curve f(x) = root(x)

- anonymous

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- anonymous

I don't get it... can anybody help please?

- mattfeury

you have to integrate on root(x) from x=0 to x=9.

- mattfeury

actually that doesn't give you a rectangle. hmmm....

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- anonymous

elaborate please

- mattfeury

this is the graph.
it is strange though because at x=0, y = 0

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- anonymous

screen shot is nothing

- anonymous

hmm, i can't see anything

- mattfeury

*hopefully*

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- mattfeury

you have to find the largest rectangle possible. looking at the graph i would guess (4,2) but that's just a though....

- anonymous

apparently we have to do it through optimization - derivatives and all that fun stuff

- anonymous

This is one of those questions you ask an instructor whose got over 20 yrs cal experience.

- mattfeury

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)

- mattfeury

try deriving that and see if find an extrema

- anonymous

not really :) as long as you have the skill then it's no prob

- anonymous

it's one of those questions that are a challenge ^_^

- anonymous

loving the enthusiasm sstarica, but i'm getting mercilessly owned in cal a

- anonymous

LOL, nah, let it be the opposite. It's quite simple, believe me :)

- anonymous

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

- anonymous

length*

- myininaya

matt has ir right! good job

- anonymous

since it's inscribed under it

- mattfeury

yes the maxima of that graph is 3!

- anonymous

:) no need for all this.

- myininaya

now we need to find A'

- anonymous

all you had to do is draw sqrt(x) and see the limit of the rectangle. ^_^

- anonymous

why A'? doesn't he want A?

- anonymous

A = 3 x 9
= 27
that if he asked for A right? ^_^

- myininaya

to maximize you find A'

- mattfeury

so it's 3 * sqrt(3)

- mattfeury

or no. 6 * sqrt(3)

- anonymous

oh right lol my bad

- myininaya

yes x=3 is right matt thats are only critcal number thats in the domain of A

- anonymous

wait wait, why is 9=xy? doesn't A=xy?

- anonymous

since he wants the largest area, then yes maximize, proceed matt and hope you understood what was going on Ire :)

- myininaya

i didnt see A=9 up there
no

- anonymous

you want the largest area, so you have to maximize in this case, they didn't ask for the "area" alone, but largest one

- mattfeury

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)

- anonymous

could you remind me how to find the maxima point please?

- anonymous

which means "optimize" lol, alight matt will take it from here ^_^ good luck

- mattfeury

take the derivative of the area function 'A'. find where the derivative = 0.

- mattfeury

normally, you'd have to do the second derivative test to find out if it is a max/min too. but i know secrets.

- anonymous

hmm

- anonymous

1. A = (9-x) * sqrt(x)

- myininaya

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- anonymous

wow, thanks guys

- mattfeury

very nice.

- myininaya

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

- myininaya

The first derivative rocks!

- anonymous

thanks for the help guys, really appreciated

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