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?
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
4 answers
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
1 answer