Pizza and Break-even Point
Complete both parts.
a) What are the three methods used for solving systems of equations? Which method do you prefer to use?
b) Break-Even Analysis – Systems of Equations Application Problem
Suppose a company produces and sells pizzas as its product. Its revenue is the money generates by selling x number of pizzas. Its cost is the cost of producing x number of pizzas.
Revenue Function: R(x) = selling price per pizza(x)
Cost Function: C(x) = fixed cost + cost per unit produced(x)
The point of intersection on a graph of each function is called the break-even point. We can also find the break-even point using the Substitution Method.
Suppose Dan’s Pizza Parlor has a fixed cost of $280 and it costs $4 to produce each pizza. Dan sells every pizza for $12.
The Revenue Function is: R(x) = 12x
The Cost Function is: C(x) = 280 + 4x
The break-even point occurs where the graphs of C and R intersect. Therefore, we can find this point by solving the system: y =12x y = 280 + 4x
How many pizzas does Dan have to produce to break-even? If he exceeds his break-even point, will he make a profit or have a loss?
A) What are the three methods used for solving systems of equations? Which method do you prefer to use? graphing, substitution, and elimination. I use all there but I think I use elimination more. B)
C(x) = 280 + 4x r(x)= 12x
12x= 280 + 4x
8x=280
x= 35 R(x) =12(35) =$420
So Dan has to produce 35 pizzas to produce his break-even point. He will have a$420 profit
a) What are the three methods used for solving systems of equations? Which method do you prefer to use?
b) Break-Even Analysis – Systems of Equations Application Problem
Suppose a company produces and sells pizzas as its product. Its revenue is the money generates by selling x number of pizzas. Its cost is the cost of producing x number of pizzas.
Revenue Function: R(x) = selling price per pizza(x)
Cost Function: C(x) = fixed cost + cost per unit produced(x)
The point of intersection on a graph of each function is called the break-even point. We can also find the break-even point using the Substitution Method.
Suppose Dan’s Pizza Parlor has a fixed cost of $280 and it costs $4 to produce each pizza. Dan sells every pizza for $12.
The Revenue Function is: R(x) = 12x
The Cost Function is: C(x) = 280 + 4x
The break-even point occurs where the graphs of C and R intersect. Therefore, we can find this point by solving the system: y =12x y = 280 + 4x
How many pizzas does Dan have to produce to break-even? If he exceeds his break-even point, will he make a profit or have a loss?
A) What are the three methods used for solving systems of equations? Which method do you prefer to use? graphing, substitution, and elimination. I use all there but I think I use elimination more. B)
C(x) = 280 + 4x r(x)= 12x
12x= 280 + 4x
8x=280
x= 35 R(x) =12(35) =$420
So Dan has to produce 35 pizzas to produce his break-even point. He will have a$420 profit