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a person wants to enclose a rectangular field with two internal partiti…

Question

a person wants to enclose a rectangular field with two internal partitions as shown. if there is 13700 feet available for fencing, what dimensions will produce the greatest area? note that the internal partitions also require fencing. the width should be feet and the length should be feet. (round your answers to two decimal places)

Explanation:

Step1: Set up the perimeter equation

Let the length be $l$ and the width be $w$. The total length of the fencing is $4w + 2l=13700$, so $l = \frac{13700 - 4w}{2}=6850 - 2w$.

Step2: Set up the area - function

The area of the rectangle $A=l\times w=(6850 - 2w)w=6850w-2w^{2}$.

Step3: Find the derivative of the area - function

The derivative $A'(w)=\frac{d}{dw}(6850w - 2w^{2})=6850 - 4w$.

Step4: Find the critical points

Set $A'(w)=0$, then $6850 - 4w = 0$. Solving for $w$ gives $4w=6850$, so $w=\frac{6850}{4}=1712.5$.

Step5: Find the length

Substitute $w = 1712.5$ into the length formula $l=6850 - 2w$. Then $l=6850-2\times1712.5=6850 - 3425 = 3425$.

Answer:

The width should be $1712.50$ feet and the length should be $3425.00$ feet.