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calc I prob optimization, correct?
I got 2029.95 but that isn't correct
Yep that's how i set it up
do you know why I subtracted 2xfrom 20 and 50?
So you do know why we subtract 2x from 20 and 50, right/
Yes i understand
So, now, we will create the function for volume. Find critical numbers Find the maximum of it (with limits we will later provide/make)
The volume would be height•lenght•width
I got 4.468
and in this case v(x)=2x•(20-2x)•(50-2x)
where V stands for volume
Why would it be 2x instead of just x
u need v`(x)
Why would it be 2x instead of just x, because you bend a cut-out of length x from both sides, don't you?
so the side 20 and side 50 is decreased by 2x. Got that, good:)
So now take the derivative of the function we got?
now we will do a regular process once we got the function V(x)=2x•(20-2x)•(50-2x)
I would perhaps simplify it a little instead of doing a product twice, but you can find the derivative as you wish:)
Oh, my bad
V(x)=x•(20-2x)•(50-2x) ^ ^ just x, not 2x.
but 20-2x and 50-2x are correct measures
Okay so take the derivative of that then?
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
V\(\rm ~'\)(x) = 1•(1000-140x+4x²)+x•(0-140+8x)
So it will be 12x^2-280x-1000?
I would use wolfram, actually, because I am the laziest person on this planet.
Hahaha i just used Wolfram for it because i didn't feel like taking the derivative myself xD
V'(x)=4(3x²-70x+250) by wolfram.... I did it like this because wolfram can give a handy version faster
set V'(x)=0 to find your critical number
remember tho, that 0>x>10 or else your volume is going to be =0 or negative.
I got 11.66 :/
that wouldn't make sense, because you get a negative volume if you plug it in our volume function.
As again 0
find the roots of 4(3x²-70x+250)=0
26.48 and -3.146 which still doesn't make sense right?
And I will tell you will get the ±, and only take the minu one, because the one with the + is greater than 10
what crtical number(s) did you get?
I am getting -60 now :// sos
take the -, because + will get x>10
35-5√19 ≈ 4.4
Oh I must have been doing some crazy math error when taking the crit #
so take 4.4 and use it in the equation at the beginning?
i got -2000 ._.
you did something wrong again-:(
Yup, that is right
Oh okay! Thank you so much of all the help. I think I understand it now!!
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