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determine the dimensions of the rectangle of maximum area that can be enclosed under these conditions

determine the dimensions of the rectangle of maximum area that can be enclosed under-example-1
User Batigolix
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1 Answer

11 votes
11 votes

The length of the pen must be less than or equal to 190ft because that's the max length of the barn.

Also, since one side of the pen doesn't need a fence, then the 280 feet of fencing will only be used by the 3 sides that is 2 width and 1 length. Therefore, 2w + l = 280ft. This our perimeter.

In terms of length, length = 280ft - 2 width.

Now, we know that the area of a rectangle is the product of length and width.

Replacing "l" with 280 - 2w in calculating the area, we have:


\begin{gathered} \text{Area}=l* w \\ \text{Area}=(280-2w)* w \\ \text{Area}=280w-2w^2 \end{gathered}

Next, let's find the vertex of this equation in which the slope is 0. To do this, let's solve for the first derivative of the equation.


\begin{gathered} slope=280-4w \\ 0=280-4w \\ 4w=280 \\ (4w)/(4)=(280)/(4) \\ w=70ft \end{gathered}

The width of the rectangular pen must be 70ft.

To solve for the length, we have:


\begin{gathered} l=280ft-2w \\ l=280ft-2(70ft) \\ l=280ft-140ft \\ l=140ft \end{gathered}

The length of the rectangular pen is 140 ft which fits to the one side of the barn that is 190ft long.

Therefore, the dimensions of the pen must be 140ft by 70ft to enclosed a maximum area of 9,800ft².

determine the dimensions of the rectangle of maximum area that can be enclosed under-example-1
User Andrew Terekhine
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