Afarm co-op wants to grow corn and soybeans. each acre of corn requires 9 gallons of fertilizer and 0.75 hours to harvest. each acre of soybeans requires 3 gallons of fertilizer and 1 hour to harvest. the co-op has available at most 40,500 gallons of fertilizer and 5,250 hours of labor for harvesting. if the profits per acre are $60 for corn and $40 for soybeans, how many acres of each should the co-op plant to maximize their profit? what is the maximum profit?

3,900 acres of corn and 3,300 acres of soybeans

Use c = corn acres and s = soybean acres. Then the hours are split by the equation:

0.5c + s = 5,250

and the fertilizer is split by the equation:

7c + 4s = 40,500

If you solve the hours for "s" and substitute into the second, you have:

7c + 4*(5,250 - 0.5c) = 40,500clear the parentheses

7c + 21,000 - 2c = 40,500subtract 21,000 from each side; simplify the left side

5c = 19,500divide each side by 5

c = 3,900

and substitute that back into the first equation:

0.5 * 3,900 + s = 5,250simplify the left side

1,950 + s = 5,250subtract 1,950 from each side

s = 3,300

you probably already have the answer but anyways it's b. 15.0

step-by-step explanation:

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: