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Calculus IChapter 4: Application of DerivativesSection 4-8

Optimization — Question 1

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Question 1

A rectangular garden is to be fenced on three sides (two widths and one length), with the fourth side along a straight river needing no fence. The area of the garden must be 120m2120 \ \text{m}^2. What dimensions will minimize the amount of fencing used?

Clearly define variables, write the function to be minimized, and find the minimum using calculus.

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Original worksheet page 1: question and worked solution for 4-8-001
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Question 1 - Solution

Let:

- xx = width (perpendicular to the river) - yy = length (along the river)

Given: - Area constraint: A=xy=120⇒y=120xA = xy = 120 \Rightarrow y = \frac{120}{x} - Fence needed: two widths and one length ⇒F=2x+y\Rightarrow F = 2x + y

Substitute into fencing function: F(x)=2x+120xF(x) = 2x + \frac{120}{x}

Minimize F(x)F(x):

Take derivative: F′(x)=2−120x2F'(x) = 2 - \frac{120}{x^2}

Set F′(x)=0F'(x) = 0: 2=120x2⇒x2=60⇒x=60=2152 = \frac{120}{x^2} \Rightarrow x^2 = 60 \Rightarrow x = \sqrt{60} = 2\sqrt{15}

Then: y=120x=120215=6015=601515=415y = \frac{120}{x} = \frac{120}{2\sqrt{15}} = \frac{60}{\sqrt{15}} = \frac{60\sqrt{15}}{15} = 4\sqrt{15}

Answer: Width = 215m\boxed{2\sqrt{15} \ \text{m}}, Length = 415m\boxed{4\sqrt{15} \ \text{m}}

Original worksheet page 2: question and worked solution for 4-8-001

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