PreCalc
posted by Ashley .
I am having a great deal of difficulty with this problem.
An open box is formed by cutting squares out of a piece of cardboard that is 16 ft by 19 ft and folding up the flaps.
a. what size corner squares should be cut to yield a box that has a volume of 175 cubic feet.
So I have this equation
x(162x)(192x)= 175
multiplied both parenthesis and got
4x^270x+304
multiplied the entire thing by x
4x^370x^2+304x=175
subtract 175
4x^370x^2+304x175=0
I'm not sure what to do next. I was thinking rational root theorem but I have tried most of the roots and it hasn't worked. I am not sure how to solve something to the third power.
Thanks

I graphed y=4x^370x^2+304x175. Perhaps that would help. I think there are a couple of solutions.

Is there a mathematical way to solve this without using the graphing calculator?

A graphical solution is already a mathematical solution. In fact, the graphical solution gives three real roots, namely near x=1, x=6 and x=11.
On the basis that 2*11>16, 11 cannot be retained as a valid solution.
So the remaining real solutions are around 1 and 6.
If you are looking for an analytical way to solve the cubic, there is the Nicolo Fontana Tartaglia method which is described in detail in:
http://www.sosmath.com/algebra/factor/fac11/fac11.html
The solution consists of three steps:
1. Find the depressed equation
by removing the x² term of the general cubic by substitution, i.e.
in f(x)=Ax³+bx²+cx+d
substitute x=yb/(3a) to give
f(y)=a*y^3 + (c(b^2)/(3*a))*y + d(b*c)/(3*a)+(2*b^3)/(27*a^2)
Note the absence of the y² term.
We will denote the depressed equation as
y³+Ay=B
2. Find s and t such that
3st=A
s²t³=B
then a real solution of the cubic is
y=st from which x can be found.
3. Using long division, reduce the cubic to a quadratic and find the two remaining solutions.
Give it a try and the three (real) roots should be around 1, 6 and 11.