We can approach the problem step by step.

Part (i): Relation for the total length of fencing wire and the area of the garden

Given:

• Sooraj’s father has 200 metres of fencing wire.
• One side of the garden is a brick wall, so no wire is required for that side.
• The other three sides (two perpendicular sides and one parallel side) require fencing.

Let:

• $x$ be the length of the side of the garden perpendicular to the brick wall.
• $y$ be the length of the side parallel to the brick wall.

The total length of the fencing wire will cover the two perpendicular sides and one parallel side: $\text{Total fencing length} = 2x + y$ We are told that the total length of fencing wire available is 200 metres: $2x + y = 200$ This is the relation representing the total length of fencing wire.

Next, the area $A(x)$ of the garden is the product of the lengths of the two sides: $A(x) = x \cdot y$ Now, using the relation $2x + y = 200$, we can express $y$ in terms of $x$: $y = 200 - 2x$ Substitute this expression for $y$ into the area formula: $A(x) = x \cdot (200 - 2x) = 200x - 2x^2$ So, the area of the garden as a function of $x$ is: $A(x) = 200x - 2x^2$

Part (ii): Determining the maximum value of $A(x)$

To find the maximum area, we need to find the value of $x$ that maximizes $A(x)$. Since $A(x) = 200x - 2x^2$ is a quadratic equation, we can use calculus or the vertex formula.

First, take the derivative of $A(x)$ with respect to $x$ to find the critical points: $A'(x) = 200 - 4x$ Set the derivative equal to zero to find the critical points: $200 - 4x = 0$ $4x = 200$ $x = 50$

Now, check the second derivative to confirm that this is a maximum: $A''(x) = -4$ Since $A''(x) = -4$, which is negative, this means $A(x)$ is concave down, and $x = 50$ is indeed the point of maximum.

Finally, substitute $x = 50$ into the equation for $y$ to find the corresponding value of $y$: $y = 200 - 2(50) = 100$

So, when $x = 50$ and $y = 100$, the area is maximized. The maximum area is: $A(50) = 50 \cdot 100 = 5000 \, \text{square metres}$

Summary:

• The relation for the total length of fencing wire is $2x + y = 200$.
• The area of the garden as a function of $x$ is $A(x) = 200x - 2x^2$.
• The maximum area of the garden is 5000 square metres when $x = 50$ and $y = 100$.

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Would you like to go over any details further, or is there anything specific you'd like clarification on?

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Here are some related questions for further practice:

1. What would happen if the length of the garden perpendicular to the brick wall was restricted to a maximum value?
2. How can we calculate the area of a garden if we know the perimeter and one dimension?
3. If the cost of the fencing material is proportional to the length, how would we minimize the cost for a fixed area?
4. What other methods could be used to optimize the area besides calculus (e.g., completing the square)?
5. How does changing the shape of the garden (e.g., making it a square) affect the area and fencing?

Tip: When dealing with quadratic functions like $A(x) = 200x - 2x^2$, the maximum or minimum value always occurs at the vertex, which can be found using the formula $x = -\frac{b}{2a}$ for a quadratic in the form $ax^2 + bx + c$.