Farmer Ed has 2,500 meters of fencing, and wants to enclose a rectangular plot that borders on a river. If Farmer Ed does not fence the side along the river, what is the largest area that can be enclosed? The largest area that can be enclosed

The largest area that can be enclosed is  781250 m²

Area of rectangle
• area = lw

where

l = length

w = width

The farmer wants to enclose a rectangular plot that borders a river. He is not fencing the side along the river. Therefore,

perimeter = l + 2w

l = 2500 - 2w

Therefore,

area = (2500 - 2w)w

(2500 - 2w)w = 0

w = 0 or 1250

average = 1250 / 2 = 625 meters

Hence, the max area is at w = 625 meters

Therefore,

l = 2500 - 2(625) = 1250

length = 1250 meters

width = 625 meter

Therefore,

area = 1250 × 625 = 781250 m²

Therefore, the largest area that can be enclosed is  781250 m²

Because it is a rectangle, the area is expressed as A = xy, or length times width.

Because it is next to the river, he only needs to fence three sides, so F = x + 2y.

Knowing the amount of fencing available is 7500m, we get:

7500 = x + 2y        solve for x

x = 7500 - 2y         substitute into the area equation

A = (7500 - 2y)y     distribute

A = -2y2 +7500y

You can see that this is a parabola which opens down, meaning that the point of maximum area will be at the vertex, y = -b/2a = -7500/[2(-2)] = 1875

x = 7500 - 2(1875) = 3750

A = 3750(1875)  = 7,031,250 m2

Step-by-step explanation:

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Find the greatest common factor of -30x^4 yz^3 and 75x^4 z^2 Fill in the blanks.

__ x^_ z^_

Hello from MrBillDoesMath!

15 x^4 z^2

Discussion:

Only x and z appear in both expressions so I am looking for the highest power of x and z in both terms. This is clearly

x^4z^2

but 30 and 75 (constant)  are both divisible by 15 so the final gcf is  15 x^4 z^2

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MrB

Town C is 2 miles east of town A and 2 miles south of Town B as in the figure at right. which of the following is the best estimate of the shortest distance from Town A to Town B?

Since C is 2mi east of A and 2mi south of B, the shortest distance from A to B would be a straight line which is equal to the hypotenuse of a right triangle with a base of 2mi and a height of 2mi..

d^2=x^2+y^2

d^2=2^2+2^2

d^2=4+4

d^2=2*4

d=√(2*4)

d=2√2 mi  (exact)

d≈2.83 mi  (to nearest hundredth of a mile)

A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through an angle Î¸. How should Î¸ be chosen so that the gutter will carry the maximum amount of water?

I = 60°

Step-by-step explanation:

Step 1: Bending one third of the sheet

So there are two segments of 10 cm on both ends that are bent upwards and a flat part of 10 cm in the center.

The more the area of the sheet the more the flow of the water. We need to find a relation between the area of the sheet and angle.

Step 2: Area of the sheet after bending can be written as

Where h is the height of the bent up segments above the base

Step 3: Relation between Area and angle I

sin I  =     ; h = 10 sin(I)

cos I =     ; b = 10 cos(I)

Therefore Area in terms of angle I can be written as

Step 4: Finding the maximum area.

To find the maximum are we need to differentiate the above equation and find the critical points of the equation

using trigonometric relation

therefore equation A'(I) becomes

Factorizing the above equation gives

To find the critical points we need to equate the above the equation to zero

Therefore,

I = 60° or 180°

Angle I can be either 60 or 180 for which the area is maximum, therefore we have to calculate the area using both of these angles to see which of them gives us the desired result.

Using

When I = 60°

When I = 180°

Therefore, the angle for which the sheet can carry the maximum water is 60° and the resulting area is 129.9 cm².

Please show how you got it...

220/2=110 each plus 34 for Mr Winkle.
Mr winkle had 134 and Mrs.Winkle had 76 which is the difference taken from her 110.

Solve the system of linear equations using the Gauss-Jordan elimination method. 2x1 − x2 + 3x3 = −10 x1 − 2x2 + x3 = −3 x1 − 5x2 + 2x3 = −7 (x1, x2, x3) =

The solution is:

Step-by-step explanation:

The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

We have the following system:

This system has the following augmented matrix:

To make the reductions easier, i am going to swap the first two lines. So

Now the matrix is:

Now we reduce the first row, doing the following operations

So, the matrix is:

Now we divide L2 by 3

So we have

Now we have:

So, now we have our row reduced matrix:

We start from the bottom line, where we have:

At second line:

At the first line

The solution is:

How to find approximate square roots

Consider the roots of the square numbers closest to the number you want the root of. From there, it's a guessing game!

A yogurt company spends \$1850 per day in fixed costs and \$0.22 for each unit. Each container of yogurt sells for \$1.25. Which function could you use to find the profit realized after selling x units of yogurt? A) f(x) = 1.47x + 1850

B) f(x) = 0.22x + 1850

C) f(x) = 1.03x − 1850

D) f(x) = 1.25x − 1850

option C

Step-by-step explanation:

given,

Yogurt company spends = \$1850 per day

cost of each unit = \$0.22

Selling price of the yogurt = \$ 1.25

the number of unit sold = x

Profit = ?

Profit = SP - C P

SP = \$1.25 x

CP = \$ 0.22 x + \$ 1850

Profit = SP - C P

Profit = \$1.25 x - (\$ 0.22 x + \$ 1850)

Profit = \$1.25 x - \$ 0.22 x - \$ 1850

Profit = \$1.03 x - \$ 1850

Hence, the correct answer is option C

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