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MArishka [77]
3 years ago
13

A rancher has 200 feet of fencing to enclose two adjacent rectangular corrals of the same dimensions. What dimensions produce a

maximum enclosed area? [Use complete-the-square to solve.]
Mathematics
1 answer:
Aloiza [94]3 years ago
7 0

Answer:

  50 ft by 33 1/3 ft

Step-by-step explanation:

Let x represent the dimension perpendicular to the shared fence. Then the dimension parallel to the shared fence will be ...

  y = (200-2x)/3

The area will be the product of these dimensions, so will be ...

  area = xy = x(200 -2x)/3 = (2/3)x(100 -x)

  area = (-2/3)(x^2 -100x)

To complete the square, we need to add the square of half the x-coefficient inside parentheses, and the opposite of that quantity outside parentheses.

  area = (-2/3)(x^2 -100x +2500) +(2/3)(2500) . . . . complete the square

  area = (-2/3)(x -50)^2 + 5000/3

The vertex of this parabolic curve is at x=50, so the dimensions of the maximum area are ...

  x = 50 . . . feet

  y = (200 -2·50)/3 = 33 1/3 . . . feet

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On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th
Elena L [17]

Answer:

The area of triangle for the given coordinates is  1.5\sqrt{4.6}

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, a = \sqrt{(3-0)^{2}+(4-0)^{2}}

Or, a =  \sqrt{9+16}

Or, a =   \sqrt{25}

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Similarly

The measure of length BC = b = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, b = \sqrt{(3-3)^{2}+(2-4)^{2}}

Or, a =  \sqrt{0+4}

Or, b =   \sqrt{4}

∴ b = 2 unit

And

So, The measure of length CA = c = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}

Or, c = \sqrt{(3-0)^{2}+(2-0)^{2}}

Or, c =  \sqrt{9+4}

Or, c =   \sqrt{13}

∴ c = \sqrt{13} unit

Now, area of Triangle written as , from Heron's formula

A = \sqrt{s\times (s-a)\times (s-b)\times (s-c)}

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I.e  s = \frac{5+2+\sqrt{13}}{2}

Or. s =  \frac{7+\sqrt{13}}{2}

So, A = \sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}

Or, A = \sqrt{(\frac{(7+\sqrt{13})}{2})\times (\frac{(\sqrt{13}-3)}{2})\times (\frac{4+\sqrt{13}}{2})\times (\frac{7-\sqrt{13}}{2})}

Or, A = \frac{3}{2} × \sqrt{1+\sqrt{13} }

∴  Area of triangle = 1.5\sqrt{4.6}

Hence The area of triangle for the given coordinates is  1.5\sqrt{4.6}  Answer

7 0
3 years ago
Read 2 more answers
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