Question

Volume and Surface Area Show that a rectangular box of given volume and minimum surface area is a cube.

Short answer

A rectangular box of given volume and minimum surface area is indeed a cube where the length of sides \(l = w = h = \sqrt[3]{\frac{V}{2}}\).

Step-by-step solution

Define the volume and surface area of a rectangular box

For a rectangular box, if length = \(l\), width = \(w\), and height = \(h\), the volume \(V\) can be calculated as \(V = lwh\), and the surface area \(A\) can be calculated as \(A = 2lw + 2wh + 2lh\).

Express the surface area equation in terms of one variable

We are given that the volume is fixed, \(V = lwh\) or we can say \(h = \frac{V}{lw}\). Substitute \(h\) from this equation into the surface area equation, we get \(A(l,w) = 2lw + 2w*\frac{V}{lw} + 2l*\frac{V}{lw}\). We now have A as function of \(l\) and \(w\). To simplify further, let's make \( l = w\) (We are trying to prove the box is a cube, which has equal sides) which turns the equation to \(A(l) = 2l^2 + \frac{2V}{l}\).

Apply calculus to find the minimum surface area

The problem now comes down to finding \(l\) such that the surface area \(A(l)\) is minimized. For that, take the derivative of \(A(l)\) with respect to \(l\), equated it to zero, then solve for \(l\). The derivative \(A'(l)\) gives us \(4l - \frac{2V}{l^2}\). Set \(A'(l) = 0\), we get \(4l = \frac{2V}{l^2}\) or \(l^3 = \frac{V}{2}\). Solving for \(l\), we get \(l = \sqrt[3]{\frac{V}{2}}\).

Verify the minimum

To verify that this is indeed a minimum and not a maximum or a point of inflection, take the second derivative. You find that \(A''(l) = 4 + \frac{4V}{l^3}\). For \(l = \sqrt[3]{\frac{V}{2}}\): \(A''(l) > 0\), this implies it's a minimum.

Conclusion

Therefore, a rectangular box of given volume with minimum surface area is a cube where the length of sides \(l = w = h = \sqrt[3]{\frac{V}{2}}\).