I have a simply proof to do, but just want to get your thoughts on it if i haven't forgot something:
I have a squared piece of metal with the side length of $ a $ and i need to cut out a open rectangular box out of it by cutting out the same quadratic edges.
My drawing of the problem (at least how i understand it):
x is in my calculation the length of one side at one of the blue squares
The definition of the volume for a rectangular box is: $$ V = a * b *c $$
since the shape we cut out the rectangular box is a square, $ a = b $ and there for:
$$ V = (a-2x) * (a-2x) * x$$
By multiplication i get:
$$ V = (a^2 - 4ax + 4x^2) * x$$ $$ V = (a^2x - 4ax^2 + 4x^3)$$
The first derivative of V is:
$$ V' = 12x^2-8ax+a^2 $$ and i need to find the zero points:
$$ V' = 12x^2-8ax+a^2 = 0 $$
therefor, using the quadratic formula i get:
$$ \frac{-b \pm \sqrt{(b^2)+4ac}}{2*a}$$
and $ a = 12 $, $ b = -8a $ and $ c = a^2 $
$$ \frac{-(-8a) \pm \sqrt{(-8a)^2+4 * 12 * a^2}}{2 * 12} $$ $$ \frac{8a \pm \sqrt{64a^2+48a^2}}{24} $$ $$ \frac{8a \pm \sqrt{16a^2}}{24} $$ $$ \frac{8a \pm 4a}{24} $$
and therefor $$ x_1 = \frac{a}{2} $$
and $$ x_2 = \frac{a}{6} $$
End of proof.
Is my proof correct or am i on the wrong path?
Thanks in advance everyone!
Overall, you are doing right steps. However, since you cut the squares from both sides of the metal sheet, the width of the box is $a-2x$. In addition, it's easier to first define the width/length of the box $y$ and then calculate the height as $(a-y)/2$. The math will be simpler.