To solve this problem, let's break down each part systematically. We are given:

• A rectangular field bordering a river, with no fence on the river side.
• The total fencing available is 1800 m.
• Let $x$ represent the width of the field (the side perpendicular to the river).

Since there is no fencing along the river, the fencing will only be used for the two widths ($2x$) and one length ($y$) of the rectangle.

(a) Expression for the length $y$ in terms of $x$

The perimeter constraint is: $2x + y = 1800$ Solving for $y$: $y = 1800 - 2x$

(b) Area of the field as a function of $x$

The area $A$ of a rectangle is given by: $A = \text{length} \times \text{width}$ Substitute $y = 1800 - 2x$: $A(x) = x \cdot (1800 - 2x)$ Expanding this: $A(x) = 1800x - 2x^2$

(c) Value of $x$ that maximizes the area

To find the maximum area, we take the derivative of $A(x)$ with respect to $x$ and set it equal to zero.

1. Differentiate $A(x) = 1800x - 2x^2$: $A'(x) = 1800 - 4x$
2. Set the derivative equal to zero: $1800 - 4x = 0$
3. Solve for $x$: $4x = 1800$ $x = 450$

Thus, $x = 450$ m leads to the maximum area.

(d) Maximum area

Now, substitute $x = 450$ back into the area function to find the maximum area: $A(450) = 1800 \times 450 - 2 \times (450)^2$ Calculating each term: $A(450) = 810000 - 2 \times 202500$ $A(450) = 810000 - 405000$ $A(450) = 405000$

Therefore, the maximum area is 405,000 square meters.

Would you like further details on any of these steps?

Related Questions

1. How would the solution change if there was fencing along all four sides?
2. What would the maximum area be if only 1000 m of fencing were available?
3. How does changing $x$ affect the area in this problem?
4. How could we confirm that $x = 450$ is a maximum using the second derivative?
5. What other shapes could maximize the area with the same perimeter constraints?

Tip

When maximizing or minimizing areas with constraints, translating all variables into a single variable can simplify differentiation and finding extremal values.