Steve has 120 feet of fence to use for making the rectangular kennel for his dogs. He uses his house as one of the sides. Let the length of the rectangle be L. The width is `(120 - L)/2` . The area of the kennel with these dimensions is `A...

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Steve has 120 feet of fence to use for making the rectangular kennel for his dogs. He uses his house as one of the sides. Let the length of the rectangle be L. The width is `(120 - L)/2` . The area of the kennel with these dimensions is `A = L*(120 - L)/2 = 60L - L^2/2`

The area is maximized when the value of L is the solution of `(dA)/(dL) = 0`

=> 60 - L = 0

=> L = 60

The width is 30 feet.

**The maximum area of the kennel is 1800 square feet.**