Calculating the dimensions of box, given volume and....

[Bekki]

Ive done it like 5 different ways, and I CANT GET IT. Im really good at these types usually, but this one is CONFUSING me.

A manufacturer produces boxes for a calculator company. The boxes have a volume of 240cm^3. Their height is 6cm less than their width, while their legnth is 1 cm less than twice their width. Find the dimensions.

So...

Legnth= (2x-1)
Width= x
Height= (x-6)

Yes?

Ok, so....

(2x-1)(x-6)(x)
gets you the cubic equation -> 2x^3-13x^2-6x


2x^3-13x^2-6x = 240

RIGHT?

OK, so thats where I get confused. Solve that equation?

 

[stapel]

2x^3-13x^2-6x = 240

OK, so thats where I get confused. Solve that equation?

First, move the 240 over to the other side of the "equals" sign. Then apply whatever tools you've learned for solving polynomial equations, such as the Rational Roots Test and synthetic division. :wink:

Have fun!

Eliz.

 

[galactus]

You have one wrong sign. Easy to do.

\(\displaystyle w(w-6)(2w-1)=240\)

\(\displaystyle 2w^{3}-13w^{2}+6w-240=0\)

Now, factor or solve by whatever means. This cubic has only one real solution. The other two are complex so you can disregard those.

To factor, try rewrting as:

\(\displaystyle (2w^{3}+3w^{2}+30w)-(16w^{2}+24w+240)\)

factor.

 

[Deleted member 4993]

I would use my graphing calculator - and find the root of the polynomial.

Then use synthetic division to confirm.

Rational root test - (240/2) has too many possibilities to be practical in this case.