Open Box Question

[Megster]

We want to make an open-topped box from 20cm by 20cm piece of cardboard by cutting out equal squares from the corners and folding flaps up to make sides. What are dimensions of each square, to nearest hundreth of a cm, so volume of resulting box would be more than 400 cm³.
I started with
400 < x (-2x+20) (-2x +20)
then eventually got
0 < x³ - 20x² + 100x - 100 (normally i would now use the factor theorem to find factors and then factor the whole equation from there and then make an interval table to find which intervals are bigger than 0 and see which work)
But i can't find any factors that make it 0 and i have no idea what to do to solve the rest. Help please and thanks

 

[soroban]

Hello, Megster!

We want to make an open-topped box from 20cm by 20cm piece of cardboard
. . by cutting out equal squares from the corners and folding flaps up to make sides.
What are dimensions of each square, to nearest hundreth of a cm,
. . so volume of resulting box would be more than 400 cm³ ?


I started with: .\(\displaystyle x(-2x+20) (-2x +20) \:>\:400\) . Yes!

then eventually got: .\(\displaystyle x^3 - 20x^2 + 100x - 100 \:>\: 0\) . Good!

Normally i would now use the factor theorem to find factors and then factor the whole equation from there,
and then make an interval table to find which intervals are bigger than 0 and see which work.
But i can't find any factors that make it 0 and i have no idea what to do to solve the rest.
Help please and thanks

Since this is Intermediate Algebra, advanced methods of solving are unavilable to us.

Note that: .\(\displaystyle 0 \,<\,x\,<\,10\)


We have a cubic function: .\(\displaystyle f(x) \:=\:x^3 - 20x^2 + 100x - 100\)
. . which has a continuous graph.

We find that: .\(\displaystyle \begin{Bmatrix}f(1) \,=\, -19 \\ f(2) \,=\, +28 \end{Bmatrix}\;\text{ and }\;\begin{Bmatrix}f(5) \,=\,+25 \\ f(6) \,=\,\;-4\end{Bmatrix}\)


We see that the function changes signs in certain intervals.

Hence, the curve crosses the \(\displaystyle x\)-axis on \(\displaystyle (1,\,2)\) and \(\displaystyle (5,\,6)\)
. . and is above the \(\displaystyle x\)-axis between the two intercepts.

I see no choice but to use trial-and-error at this point.

This is what I found . . .

. . \(\displaystyle \begin{Bmatrix}f(1.33) &\approx& -1.603 \\ f(1.34) &\approx& +0.494 \end{Bmatrix} \quad\Rightarrow\quad x \:=\:1.33\)

. . \(\displaystyle \begin{Bmatrix}f(5.87) &\approx& +0.124 \\ f(5.88) & \approx& -0.191\end{Bmatrix} \quad\Rightarrow\quad x \:=\:5.87\)


Therefore: .\(\displaystyle 1.33 \:<\:x\:<\:5.87\)

 

[Megster]

I actually figured that out last night but thank you very much! I used your answer to check over mine and make sure I was correct