Question 1098180
THE CARPENTER (OR THE CONFIDENT FIFTH-GRADER) SOLUTION:
I know how to calculate the volume of any shoe-box-shaped item.
You just multiply width, length and height.
For a cube, width length and height are all the same edge length.
If the cube edge length was 10 cm,
the initial cube volume would be
(10 cm) X (10 cm) X (10 cm) = 1000 cubic cm.
The volume of the slice cut off would be
(10 cm) X (10 cm) X (1 cm) = 100 cubic cm.
So, the final volume would be 900 cubic cm.
That is more than the 249 cm in the question.
 
I also know that the larger the original cube,
the larger the final volume,
and the larger the final volume required,
the larger the original cube needed.
 
What I know tells mne that the original cube edge must be less than 10 cm long.
It also tells me that there is only one answer.
If I calculate using increasing whole number lengths,
the calculated volumes will keep increasing,
so I will get to the answer and 249 cubic cm at some point,
or the answer was not a whole number length, and I will go past the answer.

I can calculate what the volume would be for a few edge lengths to see if I get 249 cubic cm as the answer.

If that happens, I will have the answer.
If not,
I will get less than 249 cubic cm for some whole number edge length,
but more than 249 cubic cm for the next whole number edge length,
and I will know that the answer is somewhere in between.
 
It is only common sense that the final volume is
more than the volume of a cube with edges 1 cm smaller.
I can easily calculate that a cube with edge length 5 cm has a volume of 125 cubic cm,
so the original cube's edge must be larger than 5 cm.

{{{matrix(4,5,
edge,length,5cm,6cm,highlight(7cm),
cube, volume,"125 cu.cm","216 cu.cm","343 cu.cm",
slice, volume,"25 cu.cm","36 cu.cm","49 cu.cm",
final,volume,"100 cu.cm","180 cu.cm","249 cu.cm")}}}
 
THE HIGH-SCHOOLER SOLUTION:
If
{{{x}}}= length of the edge of the cube, in cm,
{{{x^3}}}= volume of the cube in cubic cm, and
{{{x*x*1=x^2}}}= volume of a slice 1cm thick is cut from one side of the cube.
So, {{{x^3-x^2=294}}} is the volume remaining, in cubic cm.
 
All you have to do is solve {{{x^3-x^2=294}}} <--> {{{x^3-x^2-294=0}}} for {{{x}}} .
 
Solving:
{{{x^3-x^2=294}}}
{{{(x-1)x^2=294}}}
It so happens that
{{{294=2*3*7^2=6*7^2}}} , so {{{x=7}}} is a solution.
 
Is that the only solution?
The way the problwm is worded,
you would think that there is only one solution,
so answering {{{highlight(7cm)}}} ,
and going to the next problem would be a good strategy.
 
With time and willingless to spare, you could dig deeper
using whatever tools you have.
 
Using a graphing calculator, you could graph {{{f(x)=x^3-x^2-294}}} as
{{{graph(400,300,-7,9,-450,50,x^3-x^2-294)}}} and find that {{{x=7}}} is the only solution.
 
Using calculus:
{{{f(x)=x^3-x^2-294}}} could have 1 or 3 real zeros.
{{{df/dx=3x^2-2x=x(3x-2)}}} has zeros at {{{x=0}}} and {{{x=2/3}}} ,
representing respectively a local maximum and a local minimum for {{{f(x)}}} .
{{{f(0)=-294}}} is the value of {{{f(x)}}} at its local maximum,
so {{{f(x) increases for {{{x<0}}} to local maximum {{{f(0)=-294}}} ,
decreases at {{{0<x<2/3}}} to a local minimum at {{{x=2/3}}} , 
and then increases for {{{x>2/3}}} .
So, there can be only one real zero for {{{f(x)}}} ,
it happens for some {{{x>2/3}}} ,
and as we already found that {{{x=7}}} is a zero, we know know that {{{x=0}}} is the only real zero for {{{f(x)=x^3-x^2-294=0}}}.