MATH questions

using that method I got about 181 or http://www.wolframalpha.com/input/?i=%28%28%28%28220-60%28root2%29%29%5E2%29-%2860%5E2%29%29%5E.5%29%2B60 as the exact value

2x+3y+4z=24

Suppose z is fixed.

x=(24-4z-3y)/2

xyz

=(24-4z-3y)/2*yz

=(12-2z)*yz -3y^2z/2

Set derivative with respect to y to 0:

0=(12-2z)z-3yz

3yz=(12-2z)z

y=(12-2z)/3

x=(24-4z-3y)/2

=(24-4z-3((12-2z)/3))/2

=(24-4z-12+2z)/2

=6-z

Now,

xyz

=(6-z)(12-2z)z/3

=2/3 z^3 -8z^2 + 24z

set derivative to zero:

0=2z^2 -16z + 24

0=z^2-8z+12

0=(z-6)(z-2)

f(2)= 21 1/3

f(6)=0

So the maximum volume is 21 1/3, with coordinates (4, 2 2/3,2)

this is crazy

Well, I did give 2 figures, in surd notation that doesn't translate well into one line of text. It also happens to be in exact form, which I found from solving consectuively 2 quadratics.

H = 30 + 10sqrt130 + 10sqrt(130 - 6sqrt130) = 202.83...

H = 30 + 10sqrt130 - 10 sqrt(130 - 6sqrt130) = 85.205...

Put the surds in through a calculator and your irrationals appear in full glory.

I did answer the question, more exactly than in any decimal notation for irrationals. I'm sure you know of surds (specifically square roots here.) Unless the method you use couldn't come up with a neat surd form like mine? (Solving by the methods the others used in the thread, you'd need to bash out a quartic which doesn't lend well to exact answers. Clever manipulations let me solve for 2 quadratics instead.)

Right.

It's easier to use Lagrange multipliers (gradient (f(x, y)) = lamda * gradient (constraint))

Gradient of xyz = <yz, xz, xy>

Gradient of plane * lamda = <2 lamda, 3 lamda, 4 lamda>

lamda = yz/2 = xy /4

x = 2z

lamda = yz/2 = xz/3

y = 2/3 x

4z+4z+4z = 24

z = 2

x = 4

y = 8/3

Area = 2*8/3*4 = 64/3 = 21 1/3

FancyKnight - do you know if there has to be a constrain that x, y, and z are all positive, as steve bute said?

Interesting. I will read about this.

I think you do need them to be positive because you can have for instance

(x,y,z)=(-2000,8,1000)

The given solution is only the maximum when V=xyz (and not V=-xyz)

A cesaro sum does not equal a series, and you can't use algebra on one either.

And I realized my problem - if you solve for V = -xyz, you still get the same point, only as a minimum, not a maximum.

let a=b

aa=ab

aa-bb=ab-bb

(a+b)(a-b)=b(a-b)

a+b=b

b+b=b

2=1

I get something different, but its been awhile since basic algebra. Starting from this step:

(a+b)(a-b)=b(a-b)

a^2 + 2ab -b^2 = ba-b^2

2ab = 0

a=b=0

dividing by zero = infinity. should be easy to spot

655555555555+9555555555555=

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

10211111111110.

Yay!!!

Can you believe that it wasn't until I substituted a number instead of letters that I saw what it was?  That is when I realized how long it has been since school!

1800-666=1134.

turn the answer upside down on a digital calculator.

isnt that gabriels horn?

I like the pigeonhole principle too its pretty easy to comprehend

(a + b)(a - b) does not equal a^2 + 2ab - b^2.

2ab = 0 does not mean that a = b = 0.

Basically, you worked backwards and undid bradmort's work.