MATH SOLVE

7 months ago

Q:
# A farmer has a field of 70 acres in which he plants potatoes and corn. The seed for potatoes costs $20/acre, the seed for corn costs $60/acre, and the farmer has set aside $3000 to spend on seed. The profit per acre of potatoes is $150 and the profit for corn is $50 an acre. How many acres of each should the farmer plant?

Accepted Solution

A:

The farmer should plant 30 acres of potatoes and 40 acres of corn. How to determine profits in a farmIn this question we must construct the following system of three equations:Land occupied[tex]x+y = 70[/tex] (1)Net profit[tex]p = (C_{x}-c_{x})\cdot x + (C_{y}-c_{y})\cdot y[/tex] (2)Seeding costs[tex]c = c_x \cdot x + c_y \cdot y[/tex], [tex]c \le 3000[/tex] (3)Where:[tex]x[/tex] - Land occupied by potatoes, in acres.[tex]y[/tex] - Land occupied by corn, in acres. [tex]c_{x}[/tex] - Unit seeding costs for potatoes, in $ per acre.[tex]c_{y}[/tex] - Unit seeding cost for corn, in $ per acre. [tex]C_{x}[/tex] - Unit profit for potatoes, in $ per acre.[tex]C_{y}[/tex] - Unit profit for corn, in $ per acre. [tex]p[/tex] - Net profit, in $[tex]c[/tex] - Seeding costs, in $If we know that [tex]C_{x} = 150[/tex], [tex]c_{x} = 20[/tex], [tex]C_{y} = 50[/tex], [tex]c_{y} = 60[/tex], then we have the following system of equations:[tex]x + y = 70[/tex] (1)[tex]p = 130\cdot x -10\cdot y[/tex] (2b)[tex]c = 20\cdot x + 60\cdot y[/tex] (3b)By (1) in (2b):[tex]p = 130\cdot x - 10\cdot (70-x)[/tex][tex]p= 140\cdot x-700[/tex] (2c)By (1) in (3b):[tex]c = 20\cdot x + 60\cdot (70-x)[/tex][tex]c = -40\cdot x +4200[/tex] (3c)And we eliminate [tex]x[/tex] to find a relationship between profits and seeding costs:[tex]\frac{p+700}{140} = -\frac{c-4200}{40}[/tex][tex]40\cdot (p+700) = 140\cdot (4200-c)[/tex][tex]40\cdot p + 28000 = 588000-140\cdot c[/tex][tex]p = 14000 -3.5\cdot c[/tex]Linear functions do not have maxima and minima, then we must assume an arbitrary value for [tex]c[/tex] to obtain [tex]p[/tex]: ([tex]c = 3000[/tex])[tex]p = 14000 - 3.5\cdot (3000)[/tex][tex]p = 3500[/tex]By (3c):[tex]x = \frac{4200-c}{40}[/tex][tex]x = \frac{4200-3000}{40}[/tex][tex]x = 30[/tex]By (1):[tex]y = 70-x[/tex][tex]y = 40[/tex]The farmer should plant 30 acres of potatoes and 40 acres of corn. [tex]\blacksquare[/tex]To learn more on profits, we kindly invite to check this verified question: