Your revenue can be modeled by the function R(x) = 13x/x + 4.
while your cost can be modeled by the function C(x) = x + 3. Determine a fully simplified rational expression for the profit where the profit made is the difference between revenue and cost
You are responsible for organizing fundraisers for different groups. A school calls and asks you to prepare a fundraiser for them. The school prefers that you keep things simple. You want to try something new, so you think about selling fancy mathematical calculators. You do some work, enter some data into the computer and come up with two equations that will help you.
Your revenue (how much money you bring in) can be modeled by the function R(x) = 13x/x + 4 while your cost (how much you need to pay out) can be modeled by the function C(x) = x + 3. The number of calculators you sell is given by x
a) Determine a fully simplified rational expression for the profit, P(x), where the profit made is determined by the difference between revenue and cost. State your restrictions!
b) Will the school be able to sell enough calculators to break even? Be sure to provide a complete solution, including the work and the justification for your answer.
c) If your model doesn’t result in a profit, determine how you can “fix” this problem. What can you do to ensure that your model produces a profit?
For this, you might get a negative x-value as an x-intercept. What does x represent? Does it make sense to have a negative x? This is a real world problem, not just some arbitrary math problem. you can't use a discriminant if you have some wild rational function, only for when a quadratic is equal to zero. However, you may be able to rearrange P(x)=0 to show that a quadratic equals zero
For this solution, you must provide the following:
A diagram showing how the given information is used
A full solution, showing all work
Any justification/explanation that may be required
A final statement that answers the given question
R(x) = 13x/x + 4 cannot be what you mean. Parentheses matter. Surely
R(x) = 13x/(x + 4)
since profit P(x) is revenue minus cost, that makes
P(x) = 13x/(x+4) - (x+3) = (13x - (x+4)(x+3))/(x+4) = -(x^2-6x+12)/(x+4)
Pretty poor business model, since revenues level off, but costs always increase. In fact, it never shows a profit.
Now what do you have on the other parts?
hy i asked that question because i dont know but im assuming it has something to do with the answers you get for a. can you please help
what happens if C(x) = x+2 ?
no if x^2 - 6x + 9 = 0, what is the break even (WHEN this is zero) and how can we fix that
To determine the profit, we need to find the difference between revenue and cost. Let's start by subtracting the cost function C(x) from the revenue function R(x) to get the profit function P(x):
P(x) = R(x) - C(x)
Substituting the given revenue function R(x) = 13x/(x + 4) and cost function C(x) = x + 3, we have:
P(x) = 13x/(x + 4) - (x + 3)
To simplify this expression, we need to find a common denominator for the terms in the numerator of the revenue function:
P(x) = 13x - (x + 3)(x + 4)/(x + 4)
Expanding the denominator in the second term:
P(x) = 13x - (x^2 + 7x + 12)/(x + 4)
Combining like terms in the numerator:
P(x) = 13x - x^2 - 7x - 12/(x + 4)
Simplifying further:
P(x) = -x^2 + 6x - 12/(x + 4)
The fully simplified rational expression for the profit is P(x) = -x^2 + 6x - 12/(x + 4).
Now let's answer the other parts of the question.
a) The restrictions for the expression P(x) are that x cannot be equal to -4 since it would result in division by zero in the denominator. In the real-world context, it means that selling a negative number of calculators doesn't make sense.
b) To determine if the school can break even, we need to find the x-intercepts of the profit function P(x). Setting P(x) equal to zero:
0 = -x^2 + 6x - 12/(x + 4)
To simplify the equation, we multiply through by the common denominator (x + 4):
0 = (-x^2 + 6x - 12)(x + 4)
Expanding:
0 = -x^3 - 4x^2 + 6x^2 + 24x - 12x - 48
0 = -x^3 + 2x^2 + 12x - 48
Now, we can use a factoring or numerical method to find the x-intercepts. One approach is to use synthetic division or long division to find the roots. From the graph, we can see that the x-intercept is at x = 4.
c) If the profit function P(x) doesn't result in a profit, we can "fix" this problem by adjusting the selling price or reducing the cost of the calculators. Increasing the revenue or decreasing the cost will lead to a positive profit.
In conclusion, the fully simplified rational expression for the profit is P(x) = -x^2 + 6x - 12/(x + 4). The school will be able to break even if they sell enough calculators to make the profit equal to zero by finding the x-intercepts of the profit function. If the model doesn't result in a profit, adjusting the selling price or cost can help achieve a positive profit.