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Calculus IChapter 4: Application of DerivativesSection 4-8

Optimization — Question 2

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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.

  • (a) What dimensions will maximize the area of the field?

  • (b) What is the maximum area?

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Original worksheet page 1: question and worked solution for 4-8-002
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Question 2 - Solution

Let: - xx = length parallel to the river (no fence needed) - yy = width (two sides to be fenced)

From the constraint: x+2y=240⇒x=240−2yx + 2y = 240 \quad \Rightarrow \quad x = 240 - 2y

The area is: A=x⋅y=(240−2y)y=240y−2y2A = x \cdot y = (240 - 2y)y = 240y - 2y^2

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

Set derivative to zero: 240−4y=0⇒y=60240 - 4y = 0 \quad \Rightarrow \quad y = 60

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

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

(b) Maximum Area: A=120×60=7200 m2A = 120 \times 60 = \boxed{7200 \text{ m}^2}

Original worksheet page 2: question and worked solution for 4-8-002

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