<p>ok here we go....... You are planning to make an open rectangular box from a 12x14cm piece of cardboard by cutting congruent squares from the corners and folding up the sides
a) what are the dimensions of the box of largest volume you can make this way?
b)what is its volume?</p>
<p>V(x) = (14-2x)(12-2x)x</p>
<p>maximize that.. then the rest is easy</p>
<p>I remembered this problem. I guess we have the same book or something because it was in chapter 3.5 or something right? Wow, your class is slow. Mine is at Newton's Method already. Maybe we can help each other because sometimes I don't know what I'm doing either.</p>
<p>yea we move kinda sow but weve jumped around a bit i think... i slept through the first week back from vacation so i need to catch up before our test friday. I dont know what newtons method is... the prob is in chap 4.4 in my book</p>
<p>are you guys in reg calc/ab/bc?</p>
<p>I'm taking Calc AB and this is what we just went through last week.</p>
<p>i took bc last year... and i don't remember newton's method.</p>
<p>Here's</a> a mind refresher.</p>
<p>wow ur still on maximizing? And I thought my AB class was slow... We are on the mean value theorem for integrals.</p>
<p>Heh. I had that problem when I was in 5th grade (we were supposed to use guess and check). I remember asking my dad for help and him telling me that it was a pity I didn't know calc cause it would make it so much easier.</p>
<p>ok i did it and i get the dimensions of 7.72x9.72x2.14 cm</p>
<p>^^^^wow aren't you a nice person</p>
<p>do you all have the black-ish book wi/ the instrument on the front that has teh integral sign on it?</p>
<p>Let "n" equal a side of the square you have to cut out.
Length = (14-2n)
Width = (12-2n)
Height = n
Thus:
V(x) = (14-2n)(12-2n)(n) = 144n^2-24n^3.
Differentiate V(x), V'(x) = 288n - 72n^2
Set the derivative equal to zero to find maximum points.
288n - 72n^2=72(4n-n^2)
72(4n-n^2)=0
4n-n^2=0
n = {0,4}</p>
<p>0 cannot be a solution since you cannot make a box out of a flat piece of cardboard with no squares cut out. Therefore n=4 and the dimensions are:</p>
<p>6cm x 4cm x 4cm = 96cm^3</p>
<p>i have no idea what you did.. i did.
(12-2n)(14-2n)(n)=(168-24n-28n+4n^2)(n)
168n-52n^2+4n^3
f'(n)=12n^2-104n+168
then i used the quadratic equation and got n=2.14 and some other number that didnt work at that point i plugged in and got my dimensions.... so what did i do wrong??</p>
<p>can someone please tell me if i screwed up</p>
<p>i agree with you myheart, the other answer doesn't have a n^1 in the v(x) equation</p>
<p>my class already finished all the bc topics this year, we are reviewing for the ap now</p>
<p>sweet mother. i love those. i'm in pre-cal and i do those. </p>
<p>i did what lostincode did.</p>
<p>it looks easier than what hazhulken did.</p>
<p>myheartisinohio, yo uare correct. teh answers came out to 2.148 and 6.519, but as you pointed out, the later does not work as (12 - 2(6.519)) < 0</p>
<p>Whoops, I multiplied the binomials incorrectly. Sorry.
myheartisinohio, I re-did the problem and you were right. I apologize for confusing you.</p>
<p>Here's a tough one:
What is the volume of the solid of revolution obtained by rotating the region bounded by x = 1, x = 2, y = x^3, and y = 5x + 2
about the y-axis?</p>
<p>How do you find the limits of integration on this one?</p>