Posted by **Ashley** on Monday, September 7, 2009 at 1:38pm.

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(16-2x)(19-2x)= 175

multiplied both parenthesis and got

4x^2-70x+304

multiplied the entire thing by x

4x^3-70x^2+304x=175

subtract 175

4x^3-70x^2+304x-175=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

- Pre-Calc -
**bobpursley**, Monday, September 7, 2009 at 2:11pm
I graphed y=4x^3-70x^2+304x-175. Perhaps that would help. I think there are a couple of solutions.

- Pre-Calc -
**Ashley**, Monday, September 7, 2009 at 2:19pm
Is there a mathematical way to solve this without using the graphing calculator?

- Pre-Calc -
**MathMate**, Monday, September 7, 2009 at 6:26pm
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=y-b/(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

s²-t³=B

then a real solution of the cubic is

y=s-t 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.

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