


I'll give it a try, seems rather easy..
I think the first one is like this:
V=a*b*c
We know a=b
We know c=5
We know V=9245
9245=a*b*5
5ab=9245 / 5
ab=1849 (sqrt)
a=b=43
Since you cut 5 inches from each side, to form the edges for the box, the original piece has to be 10 inches wider .
43+10=53
P=53*53=2809 squareinches.
Last edited by SmileYaDead; 02292012 at 02:15 PM.

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edit: already answered >:
Last edited by Tom; 02292012 at 02:41 PM.


The second one is a real b*tch, but I'll try and think of something.


For the second number
64/4 = 16
16 for each side of the square
4 left to make the circle
I don't really understand what you need to do...


Slasher, does it guarantee the smallest total area?


I think I did the exact same problem for my precalc class last term. I'll give it a shot


Slasher is right about the numbers, we just need a decent way to show it.
Square:
P=a*b, a=b
Circle:
A=PI*r^2
r=d/2 (radius, diameter)
C/d=PI; C=d*PI; d=C/PI (circumference)
Let's say the square's side equals 16. 4*16=64; 6864=4 ;
So, C=4
Since d=C/PI, d=4/PI
r = d/2 = 4/PI/2 = 2/PI
A = PI*r^2 = PI*(2/PI)^2 = PI*(4/PI^2) = 4PI/PI^2 = 4/PI is the circle's area. Square's area=16*16=256.
Total area=256+(4/PI)=256+1.2732395447351626861510701069801=257.2732 395447351626861510701069801
To show that this is the smallest possible, we must take as big circumference of the circle as we can. So, we take 64, since we need 1 for each side of the square.
P=1*1=1
C=64
d=64/PI
r=64/PI/2=32/PI
A=PI*(32/PI)^2 = PI*(1024/PI^2) = 1024PI/PI^2=1024/PI
The total are would now be 1+1024/PI, which is about 326.94932345220164765467394738691.
Since 257. *** is smaller than 326. *** we can now see the smallest possible value of A is gotten with the smallest possible value of the circumference, which is 4.
Works?
Last edited by SmileYaDead; 02292012 at 04:21 PM.

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I think I did the exact same problem for my precalc class last term. I'll give it a shot
You have a wire that is 68 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area of the square and the circle. What is the circumference of the circle when A is a minimum?
Let l be 1 side of square.
Total length of square = 4l.
Circumference of circle = 684l.
684l=2r*pi
r=(684l)/(2pi)=(342l)/pi
Area of square = l^2
Area of circle = pi((342l)/pi)^2 = pi(4l^2136l+1156)/(pi^2)=(4l^2136l+1156)/pi
A=l^2+(4l^2136l+1156)/pi=(pi*l^2+4l^2136l+1156)/pi
Now you have two choices.
Now you have a quadratic and there are two ways of solving for the minimum length.
If you are taking calculus, you take the derivative and set it equal to 0.
As I have not taken calculus yet (taking precalc at the moment), I'll solve it without calculus.
The minimum value of A would be at the axis of symmetry.
The axis of symmetry is l=b/2a with b=136/pi and a=(4+pi)/pi.
l=136/(2(4+pi))=68/(4+pi)
Now you solve for circumference:
684l=684(68/4+pi)=68272/(4+pi)=(272+68pi272)/(4+pi)=68pi/(4+pi)
Therefore the circumference when A is minimum is 6pi/(4+pi) cm which is around 2.64cm.
This should be correct assuming I didn't mess up somewhere...
Whew that was a lot of work lol :p
Can anyone confirm/refute my solution?
Oh and calculus solution:
Derivative of A=(pi*l^2+4l^2136l+1156)/pi is (2(68+4l+pi*l))/pi
You set that to 0:
(2(68+4l+pi*l))/pi=0
136+4l+pi*l=0
4l+pi*l=136
2l=136/(4+pi)
l=68/(4+pi)
Rest is the same as other solution.
@smileyadead The total length of the square does not have to be 64 so I don't really see why that would be the solution.
Last edited by bamag; 02292012 at 05:07 PM.

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I figured it from the first one, which was rather easy and dealt with full numbers. Why would they give one that easy and the other one hard as heck? I don't really know what grade she's in tho

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