(i)[5]
Write $h$ in terms of $x$ and hence prove that $A = \frac{3}{2}x^2 + \frac{24}{x}$.
(ii)[5]
When $x$ is allowed to vary, determine the value of $x$ for which $A$ is a minimum, making it clear that $A$ is minimised rather than maximised.
Mathematics 9709 · AS & A Level · Differentiation
Write $h$ in terms of $x$ and hence prove that $A = \frac{3}{2}x^2 + \frac{24}{x}$.
When $x$ is allowed to vary, determine the value of $x$ for which $A$ is a minimum, making it clear that $A$ is minimised rather than maximised.