A rectangular allotment bed has length \(\ell\) metres and width \((\ell-3)\) metres. The area of the bed must not exceed \(40\text{ m}^2\). Show that \(\el...

This question builds a quadratic inequality modelling a maximum-area constraint, solves it by factorising, restricts it using a positivity condition on width, and then finds specific values within the resulting range before comparing the areas they give.

1. The area of the rectangular bed, length \(\ell\) and width \((\ell-3)\), is \(\ell(\ell-3) = \ell^2-3\ell\). Requiring this to be at most \(40\) m\(^2\):

   \[\ell^2-3\ell \le 40 \Rightarrow \ell^2-3\ell-40 \le 0\]

   as required [2 marks].

2. Factorising, looking for two numbers that multiply to \(-40\) and add to \(-3\), namely \(-8\) and \(5\):

   \[\ell^2-3\ell-40=(\ell-8)(\ell+5)\]

   so \((\ell-8)(\ell+5) \le 0\) gives:

   \[-5 \le \ell \le 8\]

   [2 marks]

3. Since the width \(\ell-3\) must be positive, \(\ell \gt 3\). Combining this with part (b):

   \[3 \lt \ell \le 8\]

   [1 mark]

4. The multiples of \(1.5\) strictly greater than \(3\) and at most \(8\) are shown on the number line below:

   The small open circle at \(3\) and small filled circle at \(8\) mark the boundary of the allowed range from part (c) (\(\ell=3\) excluded, \(\ell=8\) included), and the three larger filled circles mark the multiples of \(1.5\) that lie within it, giving \(\ell = 4.5\), \(\ell = 6\) and \(\ell = 7.5\) [1 mark].

5. Since the area \(\ell(\ell-3)\) increases as \(\ell\) increases throughout this range, \(\ell = 7.5\) gives the largest area:

   \[7.5 \times 4.5 = 33.75 \text{ m}^2\]

   [1 mark]

Part (c)'s condition \(\ell \gt 3\) rules out the negative part of the algebraic solution from part (b) entirely, since a length can never be negative or zero; only the physically meaningful branch \(3 \lt \ell \le 8\) survives into the final parts of the question.