A Farmer needs to enclose two different breeds of animals

Question

A farmer needs to enclose two different breads of animals in a rectangular area. The two breads are separated by a fence. There is a river on one side, so the farmer will not need any fence on that side. Therefore, the farmer will need 1 length sections of fence and 3 width sections of fence. See diagram below. The farmer has 189 ft of fence and wants the meadow to have an area of 2958 ft². What should the dimensions of the meadow be? For the purpose of this problem, the width will be the smaller dimension.

If you have two solutions for the width, use the smaller of your two answers.

If you have two solutions for the length, use the larger of your two answers.

Answer 1

Hello Jeslyn,

Hope this is useful.

Farmer has 189ft. He wants to use 3 width sections of fence. Therefore, your first equation is:

L + 3W = 189. From this equation we obtain, W=(189-L)/3

Now the area of a rectangle,

L x W = 2,958, we already know that W=(189-L)/3. Then,  

L x (189-L)/3 = 2958, simplifying we obtain,

-L^2 + 189L – 8,874 = 0, multiplying the equation by -1,

L^2 – 189L +8,874 = 0

Using the quadratic formula, we obtain to values for L,

L1= 102 and L2=87. Using these values on L +3W = 189 where, W=(189-L)/3 we obtain,

For L1=102, W1=29, and for L2=87, W2=34

Note: If you have two solutions for the width, use the smaller of your two answers.

If you have two solutions for the length, use the larger of your two answers.

Based on this, your answer is: L=102ft using 3 sections of W=29ft

 

Answer 2

This will give us two simultaneous equations in two unknown variables Length and Width:

farmer has 189 ft of fence AND river on one side, so the farmer will not need any fence on that side. The two breeds are separated by a fence.

(1) L + 3W = 189

wants the meadow to have an area of 2958 ft²

(2) LW = 2958

As the question says "What should the dimensions of the meadow be?" and doesn't specify how to solve, I will show three methods.

First and quickest, let's do a "guesstimate", by approximating to 3,000 ft², which means not so many choices for L x W:

1000 x 3

500 x 6

300 x 10

200 x 15

150 x 20

100 x 30

and we can quickly see that 100 x 30 is our best choice as 100 + 3(30) = 190

So, approximately, our field is a rectangle 100 feet x 30 feet with one long side bordered by the river, and divided into two parts by another fence of width of 30 feet.

For a more exact answer, we must rearrange (1) to make L subject, and then substitute this in (2) which gives us: L = 189 - 3W and then (189 - 3W)W = 2958

And this will give us a quadratic in W as follows:

=> -3W² + 189W = 2958

=> 3W² - 189W + 2958 = 0

=> 3(W² - 63W + 986) = 0

At this point, we could use Desmos Graphing or similar to plot the equation y = 3W² - 189W + 2958 where the x axis is W. We then read off the values where y = 0. As it's a quadratic (a parabola) we get 2 values: 29 and 34. (We are told to use the smaller of two W answers, hence 29.) And we can now find L from equation (1).

Or, we can solve the quadratic for W. Either with the quadratic formula with a = 1, b = -63 and c = 986, or, in this case, by factorizing.

3(W² - 63W + 986) = 0

=> 3(W - 29)(W - 34) = 0

=> W = 29 or W = 34 (choose the smaller as instructed)

Which we can now substitute in (1) to get

L + 3(29) = 189

=> L = 189 - 87 = 102

Hence our required rectangular field is L = 102 feet by W = 29 feet, divided into two parts by another fence of width of 29 feet.

We can double-check: L x W = 102 x 29 = 2958, as required... and L + 3W = 102 + 3(29) = 189, as required. And, it's very close to our original guesstimate.