Answer:
24cm
Explanation:
Sure, to find the height that maximizes the volume of the box, we can use calculus.
Let's denote the side length of the square base as \(x\) (since it's a square, all sides are equal), and the height of the box as \(h\). From the given condition that the sum of the dimensions is 72 cm, we have:
\[x + x + h = 72\]
\[2x + h = 72\]
\[h = 72 - 2x\]
The formula for the volume of the box is \(V = x^2 \times h\). Substituting the expression for \(h\) in terms of \(x\) into the volume equation:
\[V = x^2 \times (72 - 2x)\]
\[V = 72x^2 - 2x^3\]
To find the maximum volume, we'll take the derivative of \(V\) with respect to \(x\), set it to zero to find critical points, and then determine the \(x\) value that maximizes \(V\):
\[\frac{dV}{dx} = 144x - 6x^2\]
Setting \(\frac{dV}{dx}\) to zero to find critical points:
\[144x - 6x^2 = 0\]
\[6x(24 - x) = 0\]
So, \(x = 0\) (which doesn't make sense in this context) or \(x = 24\).
Now, let's check the endpoints and the critical point in the interval [0, 36]:
At \(x = 0\) or \(x = 36\) (the boundary values), the volume would be 0.
At \(x = 24\), which is the critical point, let's find the corresponding height:
\[h = 72 - 2x\]
\[h = 72 - 2(24)\]
\[h = 72 - 48\]
\[h = 24\]
So, when \(x = 24\), \(h = 24\), and the maximum volume is:
\[V = x^2 \times h = 24^2 \times 24 = 13824 \, \text{cm}^3\]
Therefore, to achieve the maximum volume for the cardboard box, the height should be 24 cm.