A gardener wants to build a rectangular garden of area 50 m² against an existing wall as shown in the diagram - HSC - SSCE Mathematics Advanced - Question 24 - 2023 - Paper 1

First, we need to calculate the area of the concrete path surrounding the garden. The area of the concrete path, $A$, can be defined as:

$A = 2y + x - 2$

Substituting our expression for $y$, we get:

$A = 2\left(\frac{50}{x - 2} + 1\right) + x - 2$

Simplifying:

$A = 2\left(\frac{50}{x - 2}\right) + 2 + x - 2$

This reduces to:

$A = \frac{100}{x - 2} + x$

Next, we find the derivative of $A$:

$A' = -\frac{100}{(x - 2)^2} + 1$

Setting $A' = 0$ gives:

$1 = \frac{100}{(x - 2)^2}$

This means:

$(x - 2)^2 = 100$

Solving for $x$, we get:

$x - 2 = 10 \Rightarrow x = 12$ $x - 2 = -10 \Rightarrow x = -8$

Since $x$ must be positive, we take $x = 12$.

Additionally, to confirm that this value of $x$ gives a minimum area, we evaluate $A'$ around $x = 12$. We can compute:

• For $x < 12$, $A' > 0$ (area increasing)
• For $x > 12$, $A' < 0$ (area decreasing)

This shows that there is a minimum turning point at $x = 12$.