Volume of shapes help please?

[Cassandra123]

Squares are cut down from the corners of a rectangular sheet of metal 30 inches by 50 inches. The four sides are then turned up and soldered. How large are the squares if the volume of the resulting open metal box is 232 cubic inches? How about 4000 cubic inches?

ive worked on this all day and i cant get it. please show me your steps too.

 

[lookagain]

ive worked on this all day and i cant get it. please show me your steps too.

You need to give evidence here of your work. You reach out half of the way with your work,
and then a tutor can reach out the other half of the way to you with his or her work.

 

[Cassandra123]

i usually do show my work but i really dont know where to start
ive tried guess and check like first with one inch squares then two but it all doesnt fit in.

 

[Cassandra123]

ok ijust redid it with guess and check and i got for the first bolume its 46 by 26 by 2

and fir the secind volume i got 40 by 20 by 5

but im not sure what the equation will be

 

[soroban]

Hello, Cassandra123!

Did you make a reasonable sketch?

Squares are cut down from the corners of a rectangular sheet of metal 30 inches by 50 inches.
The four sides are then turned up and soldered.
How large are the squares if the volume of the resulting open metal box is 232 cubic inches?
How about 4000 cubic inches?

The sheet of metal looks like this:

      : - - -  50 - - - - :
    - *---*-----------*---* -
    : |///:           :///| x
    : * - + - - - - - + - | -
    : |   :           :   |
   30 |   :           :   | 30-2x
    : |   :           :   |
    : * - + - - - - - + - | -
    : |///:           :///| x
    - *---*-----------*---* -
      : x :   50-2x   : x :

Those x-by-x squares are removed from each corner.
The "flaps" are folded up to form an open-top box.

The box looks like this:

          *-----------*
         /|          /| x
        / *---------/ *
       / /         / /
      *-----------* / 30-2x
    x |           |/
      *-----------*
          50-2x

\(\displaystyle \text{The volume is: }\;V \:=\:x(50\,-\,2x)(30\,-\,2x) \:=\:4x^3\,-\,160x^2\,+\,1500x\)


\(\displaystyle \text{If the volume is 4000 in}^3,\,\text{ we have: }\:4x^3\,-\,160x^2\,+\,1500x\:=\:4000\)

. . \(\displaystyle \text{which factors: }\:4(x\,-\,5)(x^2\,-\,35x\,+\,200) \:=\:0\)

. . \(\displaystyle \text{and the only real root is: }\:x\,=\,5\)

 

[Cassandra123]

i just realized i made an error. i did not mean 232 i ment 2392

and soroban how did you factor the last part out?

 

[Denis]

Soroban, x = (35 - 5SQRT(17)) / 2 is also valid ; ~7.1922