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saveliy_v [14]
3 years ago
15

A rectangular cow pasture is enclosed on three sides by a fence and the fourth side is part of the side of a barn that is $400$

feet long. The fence costs $\$5$ per foot, and $\$1,\!200$ altogether. To the nearest foot, find the length of the side parallel to the barn that will maximize the area of the pasture.

Mathematics
1 answer:
Rudiy273 years ago
3 0

Let us list out the given information:

Cost per foot = $5

Total fencing cost = $1200

Let L be the length of the barn that will maximize the area of the pasture and W be the width of the rectangular cow pasture.

From the given information, we need to do the fencing for three sides covering 1 L and 2 W's because other L will be the fourth side is part of the side of a barn. We need to write the equation based on the cost information.

If cost/foot is $5, then cost of length ' L ' will be 5L dollars and cost of '2W' width equals 10W dollars. Total cost is $1200. We can set up second equation as

5L + 10W = 1200. This can be simplified by dividing throughout the equation by 5 to get L+ 2W= 240. Solving for L we get L = 240 - 2W

Area of a rectangle is given by A = L ×W.

Substituting 240 - 2W for L we get the area function as..

A=(240 - 2W)×W= 240W-2W²

A = -2W²+240W.

When we have a function of the form f(x) = ax^2 + bx + c, then maximum value of the function can be obtained for x = -b/2a. Using that idea we can find the maximum area.

Here in the area function a = -2 and b = 240

For W = -b/2a = -240/2(-2) = -240/-4 = 60 feet, we get maximum area.

The question is asking us to find Length. We can substitute W = 60 in the equation L = 240 - 2W to find the Length.

L = 240 - 2(60) = 240 - 120 = 120 feet.

Conclusion: The length of the side parallel to the barn that will maximize the area of the pasture is 120 feet.

Note: Here the information 400 feet in the sentence "the fourth side is part of the side of a barn that is 400 feet long" given to distract us, we might not require that information to find the Length.

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2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books
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The number of ways of the books can be arranged are illustrations of permutations.

  • When the books are arranged in any order, the number of arrangements is 3628800
  • When the mathematics book must not be together, the number of arrangements is 2903040
  • When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
  • When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

\mathbf{Novels = 3}

\mathbf{Mathematics = 2}

\mathbf{Chemistry = 5}

<u />

<u>(a) The books in any order</u>

First, we calculate the total number of books

\mathbf{n = Novels + Mathematics + Chemistry}

\mathbf{n = 3 + 2 +  5}

\mathbf{n = 10}

The number of arrangement is n!:

So, we have:

\mathbf{n! = 10!}

\mathbf{n! = 3628800}

<u>(b) The mathematics book, not together</u>

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

\mathbf{Maths\ together = 2 \times 9!}

Using the complement rule, we have:

\mathbf{Maths\ not\ together = Total - Maths\ together}

This gives

\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}

\mathbf{Maths\ not\ together = 2903040}

<u>(c) The novels must be together and the chemistry books, together</u>

We have:

\mathbf{Novels = 3}

\mathbf{Chemistry = 5}

First, arrange the novels in:

\mathbf{Novels = 3!\ ways}

Next, arrange the chemistry books in:

\mathbf{Chemistry = 5!\ ways}

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

\mathbf{n =Mathematics + 1 + 1}

\mathbf{n =2 + 1 + 1}

\mathbf{n =4}

So, the number of arrangements is:

\mathbf{Arrangements = n! \times 3! \times 5!}

\mathbf{Arrangements = 4! \times 3! \times 5!}

\mathbf{Arrangements = 17280}

<u>(d) The mathematics must be together and the chemistry books, not together</u>

We have:

\mathbf{Mathematics = 2}

\mathbf{Novels = 3}

\mathbf{Chemistry = 5}

First, arrange the mathematics in:

\mathbf{Mathematics = 2!}

Literally, the number of chemistry and mathematics now is:

\mathbf{n =Chemistry + 1}

\mathbf{n =5 + 1}

\mathbf{n =6}

So, the number of arrangements of these books is:

\mathbf{Arrangements = n! \times 2!}

\mathbf{Arrangements = 6! \times 2!}

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

\mathbf{Arrangements = ^7P_3}

So, the total arrangement is:

\mathbf{Total = 6! \times 2!\times ^7P_3}

\mathbf{Total = 6! \times 2!\times 210}

\mathbf{Total = 302400}

Read more about permutations at:

brainly.com/question/1216161

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