Answer: Square base side \(x=\sqrt[3]{10}=2.15\text{ m}\), height \(h=\frac12\sqrt[3]{10}=1.08\text{ m}\).

Use differentiation to find the gradient information needed. Stationary points occur when the derivative is zero, and tangents or normals are found from the gradient at the given point.

Let the height of the tank be \(h\text{ m}\). The volume is

\(x^2h=5\), so \(h=\frac{5}{x^2}\).

Since the tank has an open top, the area of metal used is the base plus four sides:

\(A=x^2+4xh.\)

Substitute \(h=\frac{5}{x^2}\):

\(A=x^2+4x\left(\frac{5}{x^2}\right)=x^2+\frac{20}{x}.\)

Differentiate:

\(\frac{\mathrm dA}{\mathrm dx}=2x-\frac{20}{x^2}.\)

For a minimum, set this equal to zero:

\(2x-\frac{20}{x^2}=0.\)

Thus \(2x^3=20\), so \(x^3=10\), and \(x=\sqrt[3]{10}=2.154\ldots\).

Then

\(h=\frac{5}{x^2}=\frac{x}{2}=1.077\ldots\).

So the dimensions are \(2.15\text{ m}\) by \(2.15\text{ m}\) by \(1.08\text{ m}\) to 3 significant figures.

This gives the required answer: Square base side \(x=\sqrt[3]{10}=2.15\text{ m}\), height \(h=\frac12\sqrt[3]{10}=1.08\text{ m}\).