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A herpetologist is designing a rectangular experimental Iguana pen to fit in the 8' by 20' corner outside of a Komodo dragon pen. If only 42 feet of fencing is available for the Iguana pen, what dimensions should be used to obtain the largest area for the new pen? (Note: the fencing can extend, be the same length, or be shorter than the 8' wall and 20' wall, but all 42' of fencing must be used.)

Take x a the length on the 20' side and y as the length on the 8' side.

As x goes from 0' to 20', y goes from 25' down to 15'.

The perimeter is x + 2y - 8, which is always 42'.

The area goes from 0 ft² to 300 ft².

The formula for area is A = x(25 - x/2) = 25x - x²/2.

The derivative is 25 - x.

Setting this to 0 gives x = 25', which is outside the range.

This means the graph has a upwards slope across the region specified.

The maximum in this region is 300 ft², and that is at x = 20'.

As x goes from 20' to 27', y goes from 15' down to 8'.

The perimeter is 2x + 2y - 28, which is always 42'.

The area goes from 300 ft² down to 216 ft².

The formula for area is A = x(35 - x) = 35x - x².

The derivative is 35 - 2x.

Setting this to 0 gives x = 17.5', which is outside the range.

This means the graph has a downward slope across the region specified.

The maximum in this region is 300 ft², and that is at x = 20'.

There is not any point looking at the next, since the maximum is between these two.

It is at x=20', y=15', and the area is 15' x 20' = 300 ft².

Just out of curiosity, the last region was done. Here it is:

As x goes from 27' to 31', y goes from 8' down to 0'.

The perimeter is 2x + y - 20, which is always 42'.

The area goes from 216 ft² down to 0 ft².

The formula for area is A = x(62 - x) = 62x - x².

The derivative is 62 - 2x.

Setting this to 0 gives x = 31', which is outside the range.

This means the graph has a downward slope across the region specified.

The area goes from 216 down to 0, and this is less than 300,

Calculus

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