tmagloire1
  • tmagloire1
A box is to be constructed from a sheet of cardboard that is 20 cm by 50 cm by cutting out squares of length x by x from each corner and bending up the sides. What is the maximum volume this box could have? (Round your answer to two decimal places. Do not include units, for example, 10.22 cm would be 10.22.)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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SolomonZelman
  • SolomonZelman
calc I prob optimization, correct?
SolomonZelman
  • SolomonZelman
|dw:1437076558138:dw|
tmagloire1
  • tmagloire1
Yep optimization

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tmagloire1
  • tmagloire1
I got 2029.95 but that isn't correct
SolomonZelman
  • SolomonZelman
|dw:1437076685341:dw|
tmagloire1
  • tmagloire1
Yep that's how i set it up
SolomonZelman
  • SolomonZelman
do you know why I subtracted 2xfrom 20 and 50?
SolomonZelman
  • SolomonZelman
So you do know why we subtract 2x from 20 and 50, right/
tmagloire1
  • tmagloire1
Yes i understand
SolomonZelman
  • SolomonZelman
So, now, we will create the function for volume. Find critical numbers Find the maximum of it (with limits we will later provide/make)
SolomonZelman
  • SolomonZelman
The volume would be height•lenght•width
tmagloire1
  • tmagloire1
I got 4.468
SolomonZelman
  • SolomonZelman
and in this case v(x)=2x•(20-2x)•(50-2x)
SolomonZelman
  • SolomonZelman
where V stands for volume
tmagloire1
  • tmagloire1
Why would it be 2x instead of just x
SolomonZelman
  • SolomonZelman
u need v`(x)
SolomonZelman
  • SolomonZelman
Why would it be 2x instead of just x, because you bend a cut-out of length x from both sides, don't you?
SolomonZelman
  • SolomonZelman
|dw:1437076945620:dw|
tmagloire1
  • tmagloire1
ohh okay
SolomonZelman
  • SolomonZelman
so the side 20 and side 50 is decreased by 2x. Got that, good:)
tmagloire1
  • tmagloire1
So now take the derivative of the function we got?
SolomonZelman
  • SolomonZelman
now we will do a regular process once we got the function V(x)=2x•(20-2x)•(50-2x)
SolomonZelman
  • SolomonZelman
I would perhaps simplify it a little instead of doing a product twice, but you can find the derivative as you wish:)
SolomonZelman
  • SolomonZelman
Oh, my bad
SolomonZelman
  • SolomonZelman
V(x)=x•(20-2x)•(50-2x) ^ ^ just x, not 2x.
SolomonZelman
  • SolomonZelman
but 20-2x and 50-2x are correct measures
tmagloire1
  • tmagloire1
Okay so take the derivative of that then?
SolomonZelman
  • SolomonZelman
V(x)=x•(20-2x)•(50-2x) V(x)=x•(1000-40x-100x+4x²) V(x)=x•(1000-140x+4x²) I would do derivative, product now
SolomonZelman
  • SolomonZelman
V\(\rm ~'\)(x) = 1•(1000-140x+4x²)+x•(0-140+8x)
tmagloire1
  • tmagloire1
So it will be 12x^2-280x-1000?
SolomonZelman
  • SolomonZelman
I would use wolfram, actually, because I am the laziest person on this planet.
tmagloire1
  • tmagloire1
Hahaha i just used Wolfram for it because i didn't feel like taking the derivative myself xD
SolomonZelman
  • SolomonZelman
V'(x)=4(3x²-70x+250) by wolfram.... I did it like this because wolfram can give a handy version faster
SolomonZelman
  • SolomonZelman
set V'(x)=0 to find your critical number
SolomonZelman
  • SolomonZelman
numbers*
SolomonZelman
  • SolomonZelman
remember tho, that 0>x>10 or else your volume is going to be =0 or negative.
tmagloire1
  • tmagloire1
I got 11.66 :/
SolomonZelman
  • SolomonZelman
nope
SolomonZelman
  • SolomonZelman
that wouldn't make sense, because you get a negative volume if you plug it in our volume function.
SolomonZelman
  • SolomonZelman
As again 0
SolomonZelman
  • SolomonZelman
find the roots of 4(3x²-70x+250)=0
tmagloire1
  • tmagloire1
26.48 and -3.146 which still doesn't make sense right?
SolomonZelman
  • SolomonZelman
And I will tell you will get the ±, and only take the minu one, because the one with the + is greater than 10
SolomonZelman
  • SolomonZelman
what crtical number(s) did you get?
tmagloire1
  • tmagloire1
I am getting -60 now :// sos
SolomonZelman
  • SolomonZelman
|dw:1437077863587:dw|
SolomonZelman
  • SolomonZelman
take the -, because + will get x>10
SolomonZelman
  • SolomonZelman
35-5√19 ≈ 4.4
tmagloire1
  • tmagloire1
Oh I must have been doing some crazy math error when taking the crit #
tmagloire1
  • tmagloire1
so take 4.4 and use it in the equation at the beginning?
SolomonZelman
  • SolomonZelman
yes,
tmagloire1
  • tmagloire1
i got -2000 ._.
SolomonZelman
  • SolomonZelman
you did something wrong again-:(
tmagloire1
  • tmagloire1
2030.3*
SolomonZelman
  • SolomonZelman
Yup, that is right
SolomonZelman
  • SolomonZelman
2030.336
tmagloire1
  • tmagloire1
Oh okay! Thank you so much of all the help. I think I understand it now!!
SolomonZelman
  • SolomonZelman
Yes.... overview. 1. We created a function V(x)=x • (20-2x) • (50-2x) 2. Constraint x>0, because otherwise volume is negative or 0 (based on the first, x, component of V(x)). x<10, because otherwise (based on the 20-2x part), the volume would be 0 or negative. 3. We differentiated the function to get 2 critical numbers (that is two solutions of V'(x)=0) NOTE: Critical numbers (usually) are also values for which v'(x) is undefined, if defined for initial function v(x) - (but no such values in this case). Also, critical numbers usually can be closed interval boundaries, but our interval is open (we didn't include 0 or 10), so that is not our case. So in our case only solutions to v`(x)=0 are critical numbers. 4. We excluded the (35+5√19)/3 solution based on our constraint (because it is greater than 10), and rather used the only solution remaining, i.e. (35-5√19)/3. 5. We got the answr
SolomonZelman
  • SolomonZelman
yw

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