Minimum and Maximum Values — Question 9

Question 9 - Solution

(a) Express the area as a function:

Let $x$ be the length of the side perpendicular to the river. Since only three sides are fenced (two $x$’s and one along the length), the total fencing is: $$2x + y = 240 \Rightarrow y = 240 - 2x$$

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

(b) Maximize the area:

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

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

Check that this gives a maximum using second derivative: $$A''(x) = -4 < 0 \quad \Rightarrow \text{Maximum}$$

Find $y$: $$y = 240 - 2(60) = 120$$

Dimensions: $\boxed{60\, \text{m} \times 120\, \text{m}}$

(c) Maximum area: $$A = 60 \cdot 120 = \boxed{7200\, \text{m}^2}$$

Graph of $A(x) = 240x - 2x^2$: