Mathematics, 06.06.2020 10:58 crystalryan3797

A farmer needs to put in a fence in their rectangular field. If we look at the field from above, the cost of the west and east sides are $10/ft, the cost of the south side is $2/ft and the cost of the north side is $7/ft. If we have $700, use optimization to determine the dimensions of the field that will maximize the enclosed area.

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Answer from: lizarroyo

north and south sides are 38 8/9 ft longeast and west sides are 17.5 ft long

Step-by-step explanation:

Short answer: area is maximized when half the cost is spent in each of the orthogonal directions. This means the east and west sides will total $350 at $20 per foot, so will be 17.5 feet. The north and south sides will total $350 at $9 per foot, so will be 38 8/9 feet.

The dimensions that maximize the area are 17.5 ft in the north-south direction by 38 8/9 ft in the east-west direction.

Long answer: If x represents the length of the north and south sides, and y represents the length of the east and west sides, then the total cost is ...

10y +10y +2x +7x = 700

9x +20y = 700

y = (700 -9x)/20

We want to maximize the area:

A = xy = x(700 -9x)/20

We can do this by differentiating and setting the derivative to zero:

dA/dx = 700/20 -9x/10 = 0

350 -9x = 0 . . . . multiply by 10

x = 350/9 = 38 8/9

y = (700 -9(350/9))/20 = 350/20 = 17.5

The north and south sides are 38 8/9 ft long; the east and west sides are 17.5 ft long to maximize the area for the given cost.

Answer from: Quest

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2

Find the area of the regular polygon. round to the nearest tenth. !

Answer from: Quest

x= -3 and $x=\frac{-3+/-3i\sqrt{3} }{2}$

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

$(x^3-a^3)=$ $(x-a)(x^2+ax+a^2)$

where a is the cube root of $a^3$ .

here this is 27 and the cube root of 27 is 3. so we write using the form:

$(x-3)(x^2+(3)x+3^2)\\(x-3)(x^2+3x+9)$ .

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where a = 1, b=3, and c=9

$x=\frac{-3+/-\sqrt{3^2-4(1)(9)} }{2(1)} \\x=\frac{-3+/-\sqrt{9-36)} }{2} \\x=\frac{-3+/-\sqrt{-27)} }{2}\\x=\frac{-3+/-3i\sqrt{3} }{2}$

this cubic has two complex roots.

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