Trig Help Anyone

[maverick99]

I have a box with height of x.
Length of 12-2x
Width of 20-2x

And i have to show work that deterines the the answers for X that would make the volume of the box greater than 200

(20-2x)(12-2x)(x) > 200

i have been doing this for like a half hour and cant get it.

Any help would be greatly appreciated

[IslingtonFC]

Re: Trig Help Anyone

I have a box with height of x.
Length of 12-2x
Width of 20-2x

And i have to show work that deterines the the answers for X that would make the volume of the box greater than 200

(20-2x)(12-2x)(x) > 200

i have been doing this for like a half hour and cant get it.

Any help would be greatly appreciated

It's been 7 years so bare with me:

(20-2x)(12-2x)(x) > 200
20 * 12 = 240
20 * -2x = -40x
-2x * 23 = -24x
-2x * -2x = 4x squared

(240-40x-24x+4xsquared) * x > 200

240x-40xsquared - 24xsquared + 4xcubed > 200

240x - 64xsquared + 4xcubed > 200



Bare with me I'm trying to solve for X....just letting algebra come back to me.

[IslingtonFC]

Re: Trig Help Anyone

Darn dude, sorry I think you're out of luck from this point on.

I don't think you can solve for X because the equation is greater than 200...so it could be anything.

If the equation was = to 200 then you could easily bring 200 over to the other side of the equation and solve for X.

But if you use 2 as X then this happens:

240(2) - 64(2)squared + 4(2)cubed > 200
480 - 64(4) + 4(8) > 200
480 - 256 + 32 > 200
256 > 200

[maverick99]

Re: Trig Help Anyone

After Foiling and such I have to equation

(x^3 - 16x^2 + 60x + 50) = 0

There should be two answers to this at some point....because it will have to cross the Y-Axis twice....which means it will have to be equal to zero twice.

I just cannot factor that equation

[dkgojackets]

Re: Trig Help Anyone

For the record, this is not trig.

[p_rushing]

Re: Trig Help Anyone

if I remember correctly,

x^3-16x^2+60x > 50

x(x^2-16x+60) > 50
at this point, set the equation = 0
x(x^2-16x+60) = 0
x = 0 or x^2-16x+60 = 0
(x-10) = 0 (x-6) = 0
x = 10 x = 6

then you plug it back in to find the correct value, x > 10

[The_Don]

Re: Trig Help Anyone

Never mind, the post above mine has the right answer.

[dkgojackets]

Re: Trig Help Anyone

The_Don, you messed up while factoring. Dividing 240x by 4x leaves 60, not 48 as you got.

[dkgojackets]

Re: Trig Help Anyone

Here's how its done.

(20 - 2x)(12 - 2x)(x) > 200
(240 - 40x - 24x - 4x^2)x > 200
-4x^3 - 64x^2 + 240x > 200
(-4x)(x^2 + 16x - 60) > 200

Use the quadratic formula to factor the trinomial.

From graphing the equation, I'm getting 1.1738 < x < 3.8979 or x > 10.9284