Question

A wire is 1 foot long and is cut into two pieces. One piece is bent into a square and the other is bent into a semicircle. Let x= the amount used for the square and let A(x) represent the sum of the total enclosed areas of square and semicircle. Find the maximum and minimum values of A(x).

Answer 1

a negative area cannot exist however that is a verified derivative. did i go wrong or is it true that a max area would occur if he used the wire for only a semicircle?

Answer 2

my circle area is wrong...working on it

Answer 3

still obtaining a negative area ... for d/dx

Answer 4

@mathmate

Answer 5

I have not gone very far, but the area you gave for the semi-circle is actually that for the whole circle.
Can you check if that was the only problem?

Answer 6

updated

Answer 7

A(8PI/(2PI+4)=-0.0678322

Answer 8

let me see, you're good up to r=...

Answer 9

But since the area of the semi-circle is \(\pi r^2/2\), I think you forgot to square the radius.

Answer 10

fixed the negative area. but a semicircle still provides more area.

Answer 11

The A'(x) is still not up-to-date! :(

Answer 12

I got:
\(\huge A'(x)=\frac{x}{8}-\frac{\pi(1-x)}{(\pi+2)^2}\)
Hope that helps!

Answer 13

how did the square stay at the bottom?

Answer 14

The square stays at the bottom because it is a constant, so taking derivative does not affect it.

Answer 15

this?

Answer 16

Almost!
You forgot there is an x on the right hand side when you isolated x.

Answer 17

I would cross multiply and solve for x.
\((\pi+2)^2x = 8\pi (1-x)\)
expand :
\((\pi+2)^2x = 8\pi -8\pi x)\)
And
\(x = \frac{8\pi}{(\pi+2)^2+8\pi}\)
=>
\(\large x = \frac{8\pi}{\pi^2+12\pi+4}\)

Answer 18

better answer but still holds true that a semicircle maxmizes area

Answer 19

Yes, you need both max and min.

Answer 20

so the derivative is the min :)

Answer 21

You can find out if it is a max or min by finding the second derivative.
If the second derivative is >0, then it is a minimum, and if it is negative, it is a maximum.
Here the second derivative is \(\frac{\pi}{(\pi+2)^2}+\frac{1}{8}=0.24>0\), so it is a minimum.
The other two values you need to check are the boundaries, i.e. x=0, and x=1.

Answer 22

Did you calculate the area corresponding to the minimum?

Answer 23

these are tough and very involved. especially in summer school with a very very bad professor. im beginning to lose steam on this. how am i so far

Answer 24

A(0) being max and A(.24) min

Answer 25

I have xmin=8pi/.... = 0.487...
and
A(0.487...)=0.03046...
A(0)=0.0594... (semi-circle)
A(1)=0.0625 (square)

Answer 26

thank you very much for your help. i really appreciate the time youve given me. it appears though that i need to do more of these. i find it very frustrating understanding the concepts but constantly suffering from these small errors that add up and are costly and detrimental to my grade. ive spent 20-30 hrs a week on this stuff and i barely have a 70%. with only two weeks left in then summer course. with a 30 hr work week. i wish my professor didnt suck

Answer 27

Yes, I agree! Like everything else, practice makes perfect.
Summer courses are stressful, because there is less elapsed time to digest the information, and of course assignments are on your back. It is good that you spend an appreciable amount of time to work on the material. Keep working at it and you'll get your reward eventually.
One way to avoid the nuisance of costly small errors is to review your work at every step. While you're fresh at it, it takes very little time. If you have to find a mistake at the end, it will seem a monumental task.
Best of luck to your summer course, and of course, post any time!
mathmate