Optimization — Question 2

Question 2

Problem:

A rectangular field is to be enclosed on three sides with fencing (the fourth side lies along a river and requires no fence). The total length of fencing available is 240 meters.

• What dimensions will maximize the area of the field?

• What is the maximum area?

Diagram:

Question 2 - Solution

Let: - $x$ = length parallel to the river (no fence needed) - $y$ = width (two sides to be fenced)

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

The area is: $$A = x \cdot y = (240 - 2y)y = 240y - 2y^2$$

Differentiate: $$\frac{dA}{dy} = 240 - 4y$$

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

Back-substitute: $$x = 240 - 2(60) = 120$$

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

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