# Thread: Volume optimization problem

1. ## Volume optimization problem

A box with an open top is made from a square piece of cardboard, of which side length is 100 cm, by cutting a square from each corner and then folding up the sides. Find the dimensions of the box of the largest volume.

I drew a diagram and on any given side of 100 cm your going to have two pieces being cut away so I made the sum of the pieces and part left over

$\displaystyle 2x + y = 100 \Rightarrow y=100-2x \Rightarrow 2x+(100-2x) = 100$

$\displaystyle V=y^3$

and so

$\displaystyle V(x) = [100-2x]^3$ and $\displaystyle V'(x) = -6(100-2x)^2$

solving $\displaystyle V'(x)=0, x = 50$

my interval is $\displaystyle 0 \le x \le 100$

Is this correct so far? because if x = 50 than 2x=100 which doesn't seem right to me

2. ## Re: Volume optimization problem

I don't understand where you're getting a y from in the first place. If your original piece of cardboard is a square, then it has the same lengths. If you cut off x cm from each corner, that means that the length and width of your box are each going to be 100 - 2x, and the height of the box will be x.

So its volume will be \displaystyle \displaystyle \begin{align*} V = x \left( 100 - 2x \right) ^2 = x \left( 10\,000 - 400x + 4x^2 \right) = 10\,000x - 400x^2 + 4x^3 \end{align*}

This is the function you have to maximise.

3. ## Re: Volume optimization problem

Okay,

$\displaystyle V'(x) = 12x^2 -800x + 10\,000$ and solving for V'(x) = 0 I get x = 50 or x = 50/3

how do I get to the dimensions from here? The book doesn't want the volume but the dimensions that lead to the largest volume.

4. ## Re: Volume optimization problem

Well obviously if x = 50, you don't have a box because you'll have cut 100cm, so the entire side length. So x = 50/3.

If the length is 100 - 2x, and the height is x, then what are the dimensions? What is the volume?

5. ## Re: Volume optimization problem

At $\displaystyle x=0$, you have a box with height 0 centimeters. At $\displaystyle x = 50$, you have a box with base $\displaystyle (100-2\cdot 50)^2 = 0^2 = 0$ square centimeters. So, plug in $\displaystyle x = \dfrac{50}{3}$ to $\displaystyle y = 100-2x$ and you will get the length of a side of the base. $\displaystyle x$ is the height.