Loser66
  • Loser66
An open box with the length is twice its width. Find the maximum value that the volume of the box can have. Please, help.
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Loser66
  • Loser66
|dw:1438823690584:dw|
anonymous
  • anonymous
is the length 2 or width 2?
Loser66
  • Loser66
no numbers there, just that

Looking for something else?

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

More answers

anonymous
  • anonymous
2w = length of box?
Loser66
  • Loser66
yes
anonymous
  • anonymous
is the box a rectangle as stated or is it a square
Loser66
  • Loser66
My attempt: Volume of the box is V = x*W*L = x*W (2W) = 2x W^2
anonymous
  • anonymous
They said the length is 2x bigger, so w = 1 in that case
Loser66
  • Loser66
Surface are of the open box is \(S = (L+2x)(W+2x) - 4x^2= (2W+2x)(W+2x)-4x^2\) \(S = 2W^2+4xW+2xW+4x^2-4x^2= 2W^2+6xW\)
anonymous
  • anonymous
I gotta run, continue with someone else
Mertsj
  • Mertsj
How can you find the volume if you know nothing about the height of the box?
Loser66
  • Loser66
Thanks
anonymous
  • anonymous
@Mer I thought the same, anyways np loser.
Loser66
  • Loser66
@Mertsj maximum of volume at the point of derivative of the Volume =0
Loser66
  • Loser66
We consider Volume of the box is the function with respect to x and W
Loser66
  • Loser66
\(V = 2xW^2\\V'= 2W^2 + 2xW*W'\)
anonymous
  • anonymous
i think we're presuming maximum volume for a fixed perimeter, \(P=2(2x+w+2x+2w)\)
anonymous
  • anonymous
so $$P=8x+6w\implies x=\frac18(P-6w)$$
Loser66
  • Loser66
No, not that , the perimeter of what? if it is of the box, then you don't have x there, right?
Loser66
  • Loser66
|dw:1438825047689:dw| that is perimeter. One more thing, the prof asked us to work with Surface area. I am sorry for this information
anonymous
  • anonymous
fixed perimeter of the original rectangle that we cut the box out of. so now we have $$V(x,w)=w\cdot2w\cdot x=2xw^2$$now using \(x=\frac18(P-6w)\) we have: $$V=\frac14(P-6w)w^2=\frac14(Pw^2-6w^3)$$now take: $$\frac{dV}{dw}=\frac14(2Pw-18w^2)=\frac12(Pw-9w^2)$$so $$\frac{dV}{dw}=0\\Pw-9w^2=0\\w(P-9w)=0\implies w=P/9$$
Loser66
  • Loser66
oh, sorry, you are correct with perimeter. :)
anonymous
  • anonymous
which gives $$V^*=\frac14(P-6\cdot P/9)(P/9)^2=\frac14(P/3)(P^2/81)=\frac{P^3}{4\cdot 243}$$
Loser66
  • Loser66
I got what you meant. How about Surface area? how to work on it?
misty1212
  • misty1212
i get \[v(x) = 2 w^2 x-6 w x^2+4 x^3\]does that help?
anonymous
  • anonymous
and also \(x=\frac18(P-6w)=\frac18(P/3)=\frac{P}{24}\)
Loser66
  • Loser66
@misty we have to find x such that the Volume is maximum @oldrin.bataku Again, I have to work with Surface area, not perimeter. Please
misty1212
  • misty1212
yeah take the derivative wrt x get\[2 (w^2-6 w x+6 x^2)\]
misty1212
  • misty1212
set equal to zero, solve for \(x\) get something like \[(\frac{1}{2}+\frac{\sqrt3}{6})w\]
misty1212
  • misty1212
or maybe that is the wrong one
Loser66
  • Loser66
@misty1212 because V = 2xW^2 , not as your equation.
misty1212
  • misty1212
\[(\frac{1}{2}-\frac{\sqrt3}{6})w\] is probably better
misty1212
  • misty1212
hmm i get \[V=(w-2x)(2w-2x)x\] as a start
misty1212
  • misty1212
multiply out get \[V(x)=2 w^2 x-6 w x^2+4 x^3\]
Loser66
  • Loser66
@misty1212 not that, the w is the width of the box, 2w is the length of the box and x is its height
misty1212
  • misty1212
not the way you drew it for sure
misty1212
  • misty1212
oh i see then the way the question is asked, there is no maximum make w bigger and bigger the volume increases without bound
Loser66
  • Loser66
@oldrin.bataku I saw you wrote something, where is it?? why don't you post it?
anonymous
  • anonymous
so for a fixed surface area? $$A=2w^2+2wx+2(2w)x\\\quad =2w^2+6wx$$so we have \(x=\frac{A-2w^2}{6w}=\frac{A}{6w}-\frac{w}3\) so $$V=2xw^2=2\left(\frac{A}{6w}-\frac{w}3\right)w^2=\frac{Aw}{3}-\frac{2w^3}{3}\\\frac{dV}{dw}=\frac{A}3-2w^2$$so $$\frac{A}3-2w^2=0\\2w^2=A/3\\w^2=A/6\\w=\sqrt{A/6}$$
Loser66
  • Loser66
Where is your x? is it not that we have to find the ratio of x and W such that V is max? I meant like x = 1/5 W, then V is max.
anonymous
  • anonymous
plug it in: $$x=\frac{A-2\cdot A/6}{6\cdot\sqrt{A/6}}=\frac{A-A/3}{\sqrt{6A}}=\frac{2A/3}{\sqrt{6A}}=\frac{\sqrt{2A}}{3\sqrt3}$$
anonymous
  • anonymous
so the ratio of w, x is just $$x/w=\frac{\sqrt{2A}\cdot\sqrt6}{3\sqrt3\cdot\sqrt{A}}=\frac{\sqrt2\cdot\sqrt2}{3}=\frac23$$
Loser66
  • Loser66
I got exactly the same, but it is impossible when \(0
Loser66
  • Loser66
|dw:1438827425494:dw|
Loser66
  • Loser66
can't, not can
anonymous
  • anonymous
so use the extreme value theorem
anonymous
  • anonymous
if its not a local maximum in our domain, its an extreme point of our domain
Loser66
  • Loser66
Please, show me. I am struggle with it.
anonymous
  • anonymous
well, our domain has to be compact for the extreme value theorem, so we must look at \(x\in[0,w/2]\)
Loser66
  • Loser66
yes. That is the problem.
anonymous
  • anonymous
so $$x=0\implies A-2w^2=0\implies w=\sqrt{A/2}$$and $$x=w/2\implies 6w\cdot w/2=A-2w^2\implies w=\sqrt{A/5}$$
anonymous
  • anonymous
i think the problem is that these are physically unrealizable, though, and unfortunately the surface area constraint is just not strong enough
anonymous
  • anonymous
unless we've made some error
Loser66
  • Loser66
No, I went my way and work with derivative of surface area, I got exactly that x = 2/3 W plug in, V = 2x W^2 = 4/3 W ^3 let's take x = W/2, V = 2 (W/2)W^2 = W^3 < 4/3 W^3, right? testing more, we get V smaller and smaller , hence x =2/3 W seems right, but it is not right at all since it is out of domain. HA !!!
Loser66
  • Loser66
Let me show you my way \(V = 2xW^2\\V'= 2W^2+ 4xW*W'\) From \(A= 2W^2+6xW\\A'=0 = 4W*W' +6W+6xW'\) That gives \(W'=\dfrac{-6W}{4W+6x}\) replace to V' =0 \(2W^2+4xW\dfrac{-6W}{4W+6x}=0\) \(2W^2 =\dfrac{24xW^2}{4W+6x}\) \(12x = 4W +6x\\x = \dfrac{4}{6}W=\dfrac{2}{3}W\)
anonymous
  • anonymous
Actually, my way is definitely valid. the constraint to be less than $w/2$ is incorrect. It only must be that $x\ge0, w\ge0$. You are thinking of the constraint that $x$ must be less than half of the TOTAL side length $2x + w$, which is trivially true.
Loser66
  • Loser66
Algebraic way, we get stationary point is at x =(2/3)W, but it is out of the domain. I don't know how to do next.
Loser66
  • Loser66
ohhhhhhhhhhhhhh yes!!! I got you.
Loser66
  • Loser66
So, we are correct, since w is the width of the box, not the width of the paper. aaaaaaaaaaaaaahhhaaaaha... we got it right.... Thank you so so so so much.

Looking for something else?

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