A rectangular enclosure is made against a long straight wall. The wall forms one side of the enclosure and 60 m of fencing is used for the other three sides...

The wall replaces one side of the rectangle, so only three sides are fenced. That constraint links \(x\) and \(y\), which is what allows the area to be written in terms of \(x\) alone. Parts (b) and (c) then use two different techniques on the same quadratic: completing the square to find the maximum, and the quadratic formula to solve for a given value.

(a) The fencing covers the two sides of length \(x\) and the one side of length \(y\), so

\(2x + y = 60\), giving \(y = 60 - 2x\) [M1]

The area is the product of the two dimensions:

\(A = x(60 - 2x) = 60x - 2x^2\) [A1]

(b) To complete the square, first take out the factor \(-2\) from both terms containing \(x\):

\(-2(x^2 - 30x)\) [M1]

Inside the bracket, \(x^2 - 30x = (x - 15)^2 - 225\), since halving \(-30\) gives \(-15\) and \((-15)^2 = 225\) must be subtracted back:

\(-2\left[(x - 15)^2 - 225\right]\) [M1]

Multiplying out the outer \(-2\):

\(A = 450 - 2(x - 15)^2\) [A1]

A square is never negative, so \(2(x - 15)^2 \ge 0\) and the largest \(A\) occurs when the squared bracket is zero, at \(x = 15\). The greatest possible area is \(450\) m\(^2\). [A1]

(c) Set \(60x - 2x^2 = 430\) and rearrange to the standard form:

\(2x^2 - 60x + 430 = 0\), or dividing by \(2\), \(x^2 - 30x + 215 = 0\) [M1]

Using the formula with \(a = 1\), \(b = -30\), \(c = 215\), the discriminant is \(900 - 860 = 40\):

\(x = \dfrac{30 \pm \sqrt{40}}{2}\) [M1]

With \(\sqrt{40} = 6.3245\ldots\), the roots are \(\dfrac{36.3245\ldots}{2} = 18.1622\ldots\) and \(\dfrac{23.6754\ldots}{2} = 11.8377\ldots\):

\(x = 18.16\) and \(x = 11.84\) [A1]

Both roots are valid here, because both give a positive \(y = 60 - 2x\), so there are genuinely two rectangles of area \(430\) m\(^2\). They lie either side of \(x = 15\), the value that gives the maximum, which is a useful check on the completed square in part (b).