A piece of wire 6 metres long is cut into two pieces - NSC Mathematics - Question 9 - 2017 - Paper 1

The remaining length of wire used to form the rectangle BEFC is therefore:

$l = 6 - x$

Assuming that the rectangle has one side equal to the side of the square, the lengths of the rectangle can be calculated as:

$\text{Length of rectangle} = \frac{24 - 5x}{8}$

The total area (A) enclosed by the wire is the sum of the area of the square and the area of the rectangle:

$A = s^2 + l \times w$

Substituting in the expressions for the areas:

$A = \left(\frac{x}{4}\right)^2 + \left(6 - x\right) \times \frac{24 - 5x}{8}$

To find the value of x that maximizes the area, differentiate the area function A with respect to x and set the derivative to zero:

$\frac{dA}{dx} = 24 - 6x = 0$

Solving for x gives:

$x = 4$