anonymous
  • anonymous
an open box with no lid has a rectangular base. The height is equal to the shortest side of the base. What are the dimensions of the box if the volume is 208 cubic cm and the surface area is 188 square cm. **i only have two equations below that make sense. how do i start?**
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
The equations i know: vol=LWH=xxy=208 and Area = X^2+xy=188
zepdrix
  • zepdrix
Hmm let's draw it a sec.
zepdrix
  • zepdrix
|dw:1377983622808:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

zepdrix
  • zepdrix
I think your surface area equation is a little off for this one. Let's check.
zepdrix
  • zepdrix
|dw:1377983709076:dw|So we have one of these panels which has area x^2.
zepdrix
  • zepdrix
|dw:1377983760142:dw|And then we have 4 of these panels right? Which each have area xy.
zepdrix
  • zepdrix
\[\Large V=x^2y \qquad\to\qquad\quad\qquad 208=x^2y\]\[\Large A=x^2+4xy \qquad\to\qquad 188=x^2+4xy\]
zepdrix
  • zepdrix
So umm let's take our volume and write it in terms of x (solve for y). Dividing by x^2 gives us,\[\Large \color{royalblue}{y=\frac{208}{x^2}}\] We can plug this in for our y in the area equation.\[\Large 188=x^2+4x\color{royalblue}{y}\]
zepdrix
  • zepdrix
\[\Large 188=x^2+4x\cdot \color{royalblue}{\frac{208}{x^2}}\]
zepdrix
  • zepdrix
Understand what I did so far? :O Think you can solve for x from this point? :)
anonymous
  • anonymous
yes!! I realized that my surface area formula was wrong!! so the 4xy is because it has 4 faces correct?
zepdrix
  • zepdrix
ya :)
anonymous
  • anonymous
Thank you! i will check with you over the answer since I want to make sure i understand this :)
zepdrix
  • zepdrix
k sounds good.
anonymous
  • anonymous
Ok, @zepdrix since i let my width =x i solved for x by plugging in the y we found into the surface area formula. I had to factor out X^2 and i was left with (1+832x)=187. i solved for x (my width) and got 187/832 or 0.2248. does that sound correct to you?
zepdrix
  • zepdrix
\[\Large 188=x^2+4x\cdot \color{royalblue}{\frac{208}{x^2}}\]Factored out x^2? Hmm I don't think that's going to work :c\[\Large 188=x^2\left(1+x\cdot\frac{832}{x^4}\right)\]
zepdrix
  • zepdrix
The x^2 is in the denominator of that term :o factoring x^2 out `takes away` another x^2.
zepdrix
  • zepdrix
\[\Large 188=x^2+4x\cdot \color{royalblue}{\frac{208}{x^2}}\]I think what we want to do is, after simplifying this second term,\[\Large 188=x^2+\frac{832}{x}\]Multiply both sides by x and then solve the quadratic that comes out of that.
zepdrix
  • zepdrix
Oh it's not a quadratic.. hmm woops
anonymous
  • anonymous
thank you! @zepdrix I'm trying to figure it out bit by bit with your help :)
zepdrix
  • zepdrix
Hmm I must've made a boo boo somewhere. Wolfram is saying there are no real solutions to that equation. Grr one sec :d
radar
  • radar
Note that the base is rectangular, so the open top would also be rectangular (not square)
zepdrix
  • zepdrix
Ah ty! There it is! :O
zepdrix
  • zepdrix
|dw:1377985121301:dw|Two of these panels.
zepdrix
  • zepdrix
|dw:1377985137702:dw|3 of these.
anonymous
  • anonymous
ooooh so my surface area is X^2+2xy??
zepdrix
  • zepdrix
Hmm I think it works out to: \[\Large A=2x^2+3xy\]
zepdrix
  • zepdrix
We have 2 of the x^2 panels. And 3 of the (xy) panels.
anonymous
  • anonymous
I see it! I'll re-use the old method and let you and @radar know what i get. Thank you both so much! :)
zepdrix
  • zepdrix
\[\Large \color{royalblue}{y=\frac{208}{x^2}} \qquad\qquad 188=2x^2+3xy\]k :)
zepdrix
  • zepdrix
Grrr I still can't get it to work out correctly >:O @radar fix itttttttttttt :3
radar
  • radar
I'ma cipherin!. Neat problem
anonymous
  • anonymous
I don't understand this one bit >.<
zepdrix
  • zepdrix
I was able to find lengths that work by graphing it: https://www.desmos.com/calculator/xjp70ruffn But I keep messing up the algebra for some reason..
anonymous
  • anonymous
so my length is 4 cm?? I've never dealt with graphs on this kind of problem..
zepdrix
  • zepdrix
Ya graphing shouldn't be necessary for finding your x and y. I'm probably just making a silly mistake somewhere.
zepdrix
  • zepdrix
@satellite73
anonymous
  • anonymous
so far, i have 2X^2(1++312x)=188... this is my clearest step.. the rest is pretty messy.
radar
  • radar
|dw:1377986672911:dw||dw:1377987039548:dw|
radar
  • radar
Note, factoring was with help from Wolfphram :D
radar
  • radar
Messy is an apt description.
zepdrix
  • zepdrix
hmm ya :(
anonymous
  • anonymous
@rada @zepdrix you guys are my heroes! I've been stuck for about 3 hours on this problem :/ I just have one question. In the beginning, did you factor out the x^2? the solution you provided makes perfect sense but i'm trying to find understand...
radar
  • radar
Y=13 when 4 is substituted in y=208/x^2
radar
  • radar
I placed the "slimmed" down cubic equation into Wolfphram graphing engine and it gave me the three roots with the (x-4) being the rational root. You can solve the other quadratic factor and look at those other 2 irrational roots.
zepdrix
  • zepdrix
In the beginning? :o which step morn?
anonymous
  • anonymous
@radar Thank you so much! :) I'm not used to using wolfphram.. I'll definitely keep it in mind. I can't thank you or @zepdrix enough. This problem was tough!! :) i wish i could give you both medals :/ @zepdrix my question was cleared up since put the cleaned up version into wolfphram so no need to explain :) THANK YOU !!!
zepdrix
  • zepdrix
Oh ok :3
radar
  • radar
I also use @zepdrix substitution for y. I got totally hung up sustituting for x

Looking for something else?

Not the answer you are looking for? Search for more explanations.