Real world Radical Formulas
Running head: QUADRATIC FUNCTIONS
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Real World Quadratic Functions
Gail Frazier
MAT 222 Week 4 Assignment
Instructor: Simone Danielson
March 6, 2014
Real World Quadratic Functions
[no notes on this page]
-1-
QUADRATIC FUNCTIONS
2
Quadratic functions are perhaps the best example of how math concepts can be combined into a single problem. To solve these, rules for order of operations, solving equations, exponents, and radicals must be used. Because multiple variables are involved and affect the outcome, quadratics are extension of functions as well. I am to solve problem numbers 56 on page 666-667
(Dugoplski, 2012).
Maximum profit. A chain store manager has been told by the min office the daily profit, p, is
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related to the number of clerks working that day. X, according to the function p= -25x^2 +300x.
We need to figure out the number of clerks that will maximize the profit, and what the profit maximum possible profit will be.
Will this function produce a minimum or a maximize and why? How will the maximum possible profit affect the company?
-25x^2 + 300=0
divide both sides by -1
dp/dx = -50x + 300 =0
factor the left side by -1
dp/dx = -0
at maximum or maximum is -50x + 300 =0, or x= 6.
I will solve each equation
d/dx (dp/dx) or (d^2)p/ (dx^2)
is < 0 for a maximum or 0 for a minimum. The parabola will cross the x axis at 0 and 6, so (d^2) p/ (dx^2) = -50 is
1
Real World Quadratic Functions
Gail Frazier
MAT 222 Week 4 Assignment
Instructor: Simone Danielson
March 6, 2014
Real World Quadratic Functions
[no notes on this page]
-1-
QUADRATIC FUNCTIONS
2
Quadratic functions are perhaps the best example of how math concepts can be combined into a single problem. To solve these, rules for order of operations, solving equations, exponents, and radicals must be used. Because multiple variables are involved and affect the outcome, quadratics are extension of functions as well. I am to solve problem numbers 56 on page 666-667
(Dugoplski, 2012).
Maximum profit. A chain store manager has been told by the min office the daily profit, p, is
1
related to the number of clerks working that day. X, according to the function p= -25x^2 +300x.
We need to figure out the number of clerks that will maximize the profit, and what the profit maximum possible profit will be.
Will this function produce a minimum or a maximize and why? How will the maximum possible profit affect the company?
-25x^2 + 300=0
divide both sides by -1
dp/dx = -50x + 300 =0
factor the left side by -1
dp/dx = -0
at maximum or maximum is -50x + 300 =0, or x= 6.
I will solve each equation
d/dx (dp/dx) or (d^2)p/ (dx^2)
is < 0 for a maximum or 0 for a minimum. The parabola will cross the x axis at 0 and 6, so (d^2) p/ (dx^2) = -50 is