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Calculus IChapter 4: Application of DerivativesSection 4-8

Optimization — Question 4

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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.

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

  • (b) What is the maximum area that can be enclosed?

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Original worksheet page 1: question and worked solution for 4-8-004
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Question 4 - Solution

Let: - xx: length of the fence parallel to the river (only 1 needed) - yy: width of the fence perpendicular to the river (2 needed)

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

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

Differentiate: A′(y)=240−4yA'(y) = 240 - 4y

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

x=240−2(60)=120x = 240 - 2(60) = 120

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

(b) Maximum Area: A=x⋅y=120⋅60=7200m2A = x \cdot y = 120 \cdot 60 = \boxed{7200 \, \text{m}^2}

Original worksheet page 2: question and worked solution for 4-8-004

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