Well, well, well, looks like we've got some tin trouble here! To maximize the volume of this open-top box, we need some clever thinking. Let's embark on this amusing mathematical journey, shall we?
First off, we have a square sheet of tin with side length 30cm. To make an open-top box, we need to cut square corners from that sheet. The size of these squares will determine the height of the box.
Let's assume we cut a square of size 'x' from each corner. Thus, the dimensions of the resulting box will be (30 - 2x) cm by (30 - 2x) cm, with a height of 'x' cm.
So, the volume of the box can be calculated as V = (30 - 2x) * (30 - 2x) * x.
To maximize the volume, we can go down the path of differentiation! But fret not, for this will be a mild mathematics jest.
Differentiating V with respect to x, we get:
dV/dx = 4x^2 - 120x + 900.
To find the maximum (or minimum) point, we set this derivative equal to zero:
4x^2 - 120x + 900 = 0.
Solving this quadratic equation, we find two solutions, x = 7.5 and x = 15.
Now, my comedic comrade, keep in mind that x represents the size of the square we cut from each corner. Since we're dealing with a physical square sheet of tin, the cut-out squares cannot have a negative size, you see. Therefore, we'll discard the x = 15 solution.
So, after all this numerically comedic calculation, we find that cutting squares with a size of 7.5 cm from each corner will result in the box having the largest possible volume.
So grab your tin snips and get to cutting, my friend, because with this information, your open-top box will be at its volumetric zenith! Enjoy the process!