Optimization — Question 4

Question 4

Problem:

A farmer wants to build a rectangular fence next to a straight river. No fencing is needed along the river. The farmer has 240 meters of fencing available.

• What dimensions will maximize the area of the enclosure?

• What is the maximum area that can be enclosed?

Diagram:

Question 4 - Solution

Let: - $x$: length of the fence parallel to the river (only 1 needed) - $y$: width of the fence perpendicular to the river (2 needed)

Fencing constraint: $$x + 2y = 240 \quad \Rightarrow \quad x = 240 - 2y$$

Area: $$A = x \cdot y = (240 - 2y)y = 240y - 2y^2$$

Differentiate: $$A'(y) = 240 - 4y$$

Set derivative to 0: $$240 - 4y = 0 \quad \Rightarrow \quad y = 60$$

$$x = 240 - 2(60) = 120$$

(a) Optimal Dimensions: $$\boxed{x = 120 \, \text{m}, \quad y = 60 \, \text{m}}$$

(b) Maximum Area: $$A = x \cdot y = 120 \cdot 60 = \boxed{7200 \, \text{m}^2}$$