Question

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)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

elaborate please

Answer 5

this is the graph.

it is strange though because at x=0, y = 0

Answer 6

screen shot is nothing

Answer 7

hmm, i can't see anything

Answer 8

*hopefully*

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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)

Answer 13

try deriving that and see if find an extrema

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

length*

Answer 20

matt has ir right! good job

Answer 21

since it's inscribed under it

Answer 22

yes the maxima of that graph is 3!

Answer 23

:) no need for all this.

Answer 24

now we need to find A'

Answer 25

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

Answer 26

why A'? doesn't he want A?

Answer 27

A = 3 x 9
= 27

that if he asked for A right? ^_^

Answer 28

to maximize you find A'

Answer 29

so it's 3 * sqrt(3)

Answer 30

or no. 6 * sqrt(3)

Answer 31

oh right lol my bad

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

i didnt see A=9 up there
no

Answer 36

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

Answer 37

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)

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

hmm

Answer 43

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

Answer 44

wow, thanks guys

Answer 45

very nice.

Answer 46

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

Answer 47

The first derivative rocks!

Answer 48

thanks for the help guys, really appreciated