Question

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Answer 1

something isn't quite right here I think... those are points on the base of the cone.

Answer 2

z can just equal its constraint of 1? any value less than that would decrease the volume obviously

Answer 3

you should be looking at boxes like:

Answer 4

otherwise your box is only going to be flat... (not a box)

Answer 5

i see your point, i just dont see where i couldve gone wrong though solving through lagrange, i just took the partials of f(x,y,z) = xyz and g(x,y), then solved the two simaltaneous equations. hmmm.

Answer 6

I think you just used the base of the cone as your constraint, rather than the cone itself.. right?

Answer 7

that won't work:)

Answer 8

the constraint for this problem is that x y and z for the box must lie on the surface of the cone...

Answer 9

hmmm, so i need an equation for a plane with height of 1 for the cone?

Answer 10

elliptical cones are discussed:
http://tutorial.math.lamar.edu/Classes/CalcIII/QuadricSurfaces.aspx

Answer 11

you're going to have to do this problem upside I think :)

Answer 12

thankyou! so all i need to do is find the equation for the cone tapering to (0,0,1) and do lagrange through that?

Answer 13

starting at z=1 to z=0

Answer 14

what are the a, b and c values for in the cone equation? is that something to do with specifying for example that z goes from 1 to 0?

Answer 15

when z=1
x^2+4y^2=4
that should be enough...?

Answer 16

sorry stepped away for a bit.

Answer 17

mm all i can get from that is radius is 2? really not sure how to get the equation of the surface, sorry.

Answer 18

radius isnt constant just realised

Answer 19

think it's x^2 + 4y^2 = 4z^2

Answer 20

is that taken from the general formula at the bottom of the link you posted? thanks for your help :)