Post a New Question

Pre-Calc

posted by on .

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 - ,

    I graphed y=4x^3-70x^2+304x-175. Perhaps that would help. I think there are a couple of solutions.

  • Pre-Calc - ,

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

  • Pre-Calc - ,

    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.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question