A homeowner has 60 feet of fencing material to enclose a rectangular area for his pets to play in. He will use one side of his house as a side of the play area.  What dimensions should he use if...

A homeowner has 60 feet of fencing material to enclose a rectangular area for his pets to play in. He will use one side of his house as a side of the play area.  What dimensions should he use if he want to maximize the play area.

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The homeowner has 60 feet of fencing material to enclose a rectangular area for his pets to play in. One side of the play area would be his house. The dimensions of the play area have to be determined to maximize it.

Let the length of the side that would be the house be L. As the total length of fencing material is 60, if the other side is x, 2x + L = 60

=> L = 60 - 2x

The area of the house is A = x*(60 - 2x) = 60x - 2x^2

Differentiate A with respect to x, equate it to 0 and solve the equation for x.

`(dA)/(dx) = 60 - 4x`

60 - 4x = 0

=> x = 15

One of the sides of the area is 15 and the other is 30 feet.

To maximize the area, the dimensions of the play area are 15 feet and 30 feet.

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taylorruth's profile pic

taylorruth | (Level 1) Salutatorian

Posted on

If the area of a rectangle is A=lw , where does the division come into this answer?

nisarg's profile pic

nisarg | Student, Grade 11 | (Level 1) Valedictorian

Posted on

since one side of the fence will be the house he could use 15 feet by 30 feet because that would make an area of 450ft squared

ayl0124's profile pic

ayl0124 | Student, Grade 12 | (Level 1) Valedictorian

Posted on

The area of a rectangle is:

`A = lw`

You have a restriction:

`60 = 2l + w`

Solve for one of the variables.

`w = 60 -2l`

Your new area equation should be:

`A = l(60-2l)`

or

`A = 60l- 2l^2`

Find the derivative of this equation.

`A' = (60 - 4l) "dl"`

One important factor to know about calculus is that when the derivative of an equation equals zero (changes signs), there is a maximum or minimum. 

`0 = 60 - 4l`

Solve for "l":

`60 = 4l`

`l = 15`

Using your initial equation, you can now solve for "w."

`60 = 2(15) + w`

`w = 30`

Therefore, your dimensions should be 15 x 30, with two sides equalling 15 and one side equaling 30.

rachellopez's profile pic

rachellopez | Student, Grade 12 | (Level 1) Valedictorian

Posted on

First off, you know that the space is a rectangle, which generally have two short sides of equal length and two long sides of equal length. Often enough the long side will be double the short side.

However, you want to assume you will use all 60 feet of fencing. If you had two short sides that were 10 feet and a long side that was 20 feet you'd only be using 40 feet of fencing. If your short sides were 20 feet, your long side would be 40 and that would mean you'd need 80 feet of fencing, which you don't have enough for. To make you rectangle big enough with the amount of fencing you'd need your short sides to be 15 feet each and your long side 30 feet. This uses your 60 feet of fencing and creates a big space for the animals.

I always just use logic for these problems; I was never really given a formula. This "guess and check" method is easier to do with multiple choice options, but it is just an alternative way to think about it.

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