#extreme values, critical numbers, concaves, related rates
78 messages · Page 1 of 1 (latest)
Its optimization, they have pretty much the same methods of solving, it just varies if maximum or minimum is being asked. You can look up on yt for optimization problem and try
what about the first one
Do you know what the slope of an extreme value is?
solve for critical numbers by finding the first derivative then setting it equal to 0
analyze concavity inflectivity yada yada by finding the second derivative then factoring
does the volune have to do with anything on question 1? if not, use AM-GM with 3 numbers (a^3+b^3+c^3 >= 3abc for a;b;c >= 0 with equality happening at a=b=c or a+b+c = 0 (which can't happen in this situation), try proving that first, then try to think of how you would use it to get rid of the x)
equality happen at x^3 = 50 or x = cuberoot(50) which is in the [1;10] range
could you explain how you got to this solution?
AM-GM with 3 numbers
a^3 + b^3 + c^3 >= 3abc for a;b;c>=0
don't just give away the answer, guide the OP
okay i'll edit it ig
also, i'm pretty sure the volume is important, as it belongs to question 1
still don't know why the volume is in there tho maybe it's a red herring
square base and a volume of 50 ft³
is it just a box with all 6 sides being a square?
i'm thinking of it as a rectangle box so the height could be anything
i think so too
so the 2 sides on top and bottom are square and the 4 other sides are rectangle?
otherwise the problem would be very easy
i'll try that ig
as then length = width = height
faces*
if it's this then call the height h, the side of the square x
volume = 50 ft^3 or x^2h = 50 ft^3
so surface area = 4xh
4xh = 2x^2 + 200/x
4x^2h = 2x^3 + 200
200 = 2x^3 + 200
2x^3 = 0 or x = 0
if that was the case then x = 0 which won't make any sense, so i'm really just assuming the volume is a red herring
oh im assuming S(x) = 2x^2 + 200/x where x is in term of ft,it didn't mention in the question but since the volume is measured in ft x should also be in ft i think
the question asks for the minimum surface area on the interval [1, 10]
@sterile thorn yeah i did solve it like this, once you get x you can just solve for h using x^2h = 50
i'm just wondering why they list the volume in the first place
is the surface area the area of 6 sides then?
i think to verify your answer
it's the total surface area of all faces
oh okay then, in my country there's a difference between the area of the 4 faces around all all 6 faces, i'll do it with 6 faces i guess
in that case, the area of the base would also be 0
you take the derivative of the surface area formula and find the x value where the rate of change is 0
because thats where the curve is level
meaning its the bottom or top point of the function
so that's where the extreme values are
i'm pretty sure this way to solve is actually correct atleast to my interpertation of the question, here's my solution
|| let's call the side of the square x (ft), and the height h (ft)
volume = 50 ft^3, therefore x^2h = 50 ft^3
surface area = 2x^2 + 4xh
S(x) is the surface area, therefore
2x^2 + 4xh = 2x^3 + 200/x
and x is not 0, so 2x^3 + 4x^2h = 2x^3 + 200, plug x^2h = 50ft^3 we have
2x^3 + 200 = 2x^3 + 200 which is always true, so we just have to calculate the minimum of 2x^2 + 200/x and find the value of x where equality happen then solve for h
with a;b;c >= 0, a^3+b^3+c^3 - 3abc = 1/2 (a+b+c)*((a-b)^2+(b-c)^2+(c-a)^2) >= 0 therefore a^3 + b^3 + c^3 >= 3abc or a +b + c >= 3cuberoot(abc)
apply to the question we have, 2x^2 + 200/x = 2x^2 + 100/x + 100/x >= 3cuberoot(20000)
equality happens when 2x^2 = 100/x or x^3 = 50
x = cuberoot(50) ft which is in the [1;10] range so it is valid
h = cuberoot(50) ft ||
local tops and bottoms
my interpertation is that the box is this shape
that's my interpretation as well
the question asks for the minimum, so ImOakley is right
well yeah what i did also get the minimum
as the other questions are also about extreme values
well here's the answer i get i guess, you guys can check if it's the same
Absolute minimum || 3cuberoot(20000) ||
Dimension where absolute minimum happens || The box is a cube with a side length of cuberoot(50) ||
you still need to simplify
simplify at what part?
okay then
Absolute minimum ||30*cuberoot(20)||
Dimension where minimum happens ||The box is a cupe with a side length of cuberoot(50)||
the minimum has to be on the interval [1, 10]
according to my calculator, your solution doesn't
it's too big
oh alright, i thought that's the range for x not the whole thing
good point, now i'm not sure
idk how it can be from 0 to 10 though, the value i calculated is already the lowest so im not sure how to go lower
maybe it can with negatives?
how can a length or volume be negative
yeah idk, if everything is positive then the lowest value should def not be below 10 unless the [1;10] measure isn't ft^2
the minimum value of 2x^2 + 200/x is already way larger than 10 for all x > 0 and there's also more conditions on top of that (tbf in my solution the volume doesn't affect anything as you can still find a valid value of the height where it works)
yup, the vertical asymptote is x = 0
nevermind, you are correct
but the solutions are in ft^2 and ft
can't you subtract 2x^2 from both sides?
Alright then
Yeah I'm just proving S(x) is always equal to the surface area
because it's the formula for the surface area
but you can subtract the base area term from both sides
i think
Yeah but I need to prove S(x) is always gonna be equal to the surface area first so I can make sure that it's possible to find h once I find the equality for min S(x)
but the question states that it is the formula for the surface area
Okay then I just wanna make sure
If it's smt like S(x) = x^2 + 250/x it'll only be true for a certain value of x