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close a rectangular field with five internal partitions as shown. if th…

Question

close a rectangular field with five internal partitions as shown. if there is 16200 feet, what dimensions will produce the greatest area? note that the internal partitions also

Explanation:

Step1: Define variables

Let the length of the rectangle be $x$ and the width be $y$. The total length of the fencing used is $6y + 2x$ (6 internal - partitions along the width and 2 lengths). Given that the total length of the fencing is 16200 feet, so $6y+2x = 16200$. We can rewrite this equation as $x=8100 - 3y$.

Step2: Express the area function

The area of the rectangle $A=xy$. Substitute $x = 8100 - 3y$ into the area formula, we get $A(y)=(8100 - 3y)y=8100y-3y^{2}$.

Step3: Find the derivative of the area function

Differentiate $A(y)$ with respect to $y$. $A^\prime(y)=\frac{d}{dy}(8100y - 3y^{2})=8100-6y$.

Step4: Find the critical points

Set $A^\prime(y)=0$. Then $8100 - 6y = 0$. Solving for $y$ gives $6y=8100$, so $y = 1350$.

Step5: Find the second - derivative of the area function

Differentiate $A^\prime(y)$ with respect to $y$. $A^{\prime\prime}(y)=\frac{d}{dy}(8100 - 6y)=-6<0$. Since $A^{\prime\prime}(y)<0$ when $y = 1350$, the area function has a maximum at $y = 1350$.

Step6: Find the value of $x$

Substitute $y = 1350$ into the equation $x=8100 - 3y$. Then $x=8100-3\times1350=8100 - 4050 = 4050$.

Answer:

The width should be 1350.00 feet and the length should be 4050.00 feet.