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klio [65]
3 years ago
10

The owner of the Rancho Grande has 3,060 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 (in yd) of the largest area he can enclose?
Mathematics
1 answer:
Pavel [41]3 years ago
3 0

Answer:

1170450 yd^2

Step-by-step explanation:

The first thing is to calculate the necessary perimeter, which would be like this:

2 * a + b = 3060

if we solve for b, we are left with:

b = 3060-2 * a

Now for the area it would be:

A = a * b = a * (3060-2 * a )

A = 3060 * a -2 * a ^ 2

To maximize the area, we calculate the derivative with respect to "a":

dA / da = d [3060 * a -2 * a ^ 2

]/gives

dA / day = 3060 - 4 * a

If we equal 0:

0 = 3060 - 4 * a

4 * a = 3060

a = 3060/4

a = 765 and d

Therefore b:

b = 3060 - 2 * a = 3060 - 1530 = 1530

A = a * b

A = 765 * 1530

A = 1170450 and d ^ 2

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jekas [21]

Answer:

y = -5x - 21

Step-by-step explanation:

Given in the question,

equation of a parallel line

y = -5x + 6

point through which it passes

(-4,-1)

Step1

Find the gradient of the equation given, as it is parallel so it will have same gradient

equation of straight line

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Step2

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-1 = -5(-4) + c

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c = -20 - 1

c = -21

Step3

form the equation

y = -5x - 21

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