The owner of the Rancho Grande has 3020 yd of fencing with which to enclose a rectangular piece of grazing land situated along the straight portion of a river. If fencing is not required along the river, what are the dimensions of the largest area he can enclose?shorter side yd
longer side yd

What is this area?
yd2

Answers

Answer 1
Answer:

Answer:

Shorter side = 755 yd

longer side   = 1510 yd  

A(max)  = 1140050 yd²

Step-by-step explanation:

The owner has 3020 yd of fencing

Lets assume:

x the shorter side of the rectangle ( we will use fencing in two sides of length x )

y the longer one

A area of the rectangle          and P the perimeter of the rectangle ( we have only three sides covered by fencing material)

We have:       A = x * y          P = 2 * x + y   ⇒  y  = P - 2*x  ⇒  y  = 3020 - 2* x

A (x) = x * ( 3020 - 2*x)  ⇒  A(x) = 3020 * x - 2* x²

Taken derivative

A´(x) = 3020 - 4 * x

If     A´(x) = 0               3020 -4*x = 0     ⇒4*x = 3020    x =  755 yd

If we take second derivative A´´(x)  = -4

so  A´´(x) < 0 so there is a maximun in point x = 755

Then

Rectangle dimensions :

x = 755 yd        ⇒  y = 3020 - 2 * x   ⇒ y = 3020 - 2 * (755)     y = 1510 yd

Maximum area is :  A(max)  = 1510 * 755  ⇒ A(max)  = 1140050 yd²

Answer 2
Answer:

Final answer:

To get the largest area, 3020 yards of fencing is divided equally to form a rectangular grazing piece. Each side is 1510 yards, leading to a total area of 2,280,100 square yards.

Explanation:

We are given a total of 3020 yards of fencing which is used to fence two sides of a rectangle, with a river enclosing the third side. The largest area of a rectangle is obtained when the rectangle is a square. However, since one of the sides is the river, and hence not fenced, the rectangle is not square but should be as close to a square as possible to give the maximum area.

The optimum distribution is dividing the fence into two equal parts for both sides of the rectangle, so each side will be 3020 / 2 = 1510 yds. The dimensions of the largest area he can enclose are: shorter side = yd, longer side = 1510 yds.

The area of this rectangular grazing land would then be 1510 yd * 1510 yd = 2,280,100 yd2.

Learn more about Maximum Area here:

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Answers

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Answers

Answer:

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Step-by-step explanation:

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Todd and Garrett began arguing about who did better on their tests, but they couldn't decide who did better given that they took different tests. Todd took a test in Math and earned a 74.6, and Garrett took a test in English and earned a 68.8. Use the fact that all the students' test grades in the Math class had a mean of 70.6 and a standard deviation of 11.9, and all the students' test grades in English had a mean of 63.7 and a standard deviation of 8.6 to answer the following questions. Required:
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Answers

Part A

For Todd's class, we have this given info:

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  • sigma = 11.9 = population standard deviation of math scores

Compute the z score for x = 74.6

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Side note: Convention usually has z scores rounded to two decimal places. If your teacher instructs otherwise, then go with those instructions of course.

Answer:  0.34

======================================================

Part B

For Garret's English class, we have:

  • mu = 63.7
  • sigma = 8.6

Compute the z score for x = 68.8

z = (x-mu)/sigma

z = (68.8 - 63.7)/(8.6)

z = 0.59 approximately

Answer:  0.59

======================================================

Part C

Garret has the higher z score, which means that Garret did relatively better to his classmates compared to Todd's performance (in relation to his classmates). The z score is the distance, in units of standard deviation, the score is from the mean. Positive z values are above the mean, while negative z values are below the mean.

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Answers

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

hope i helped hehe :)