SOLUTION: For a certain company, the cost for producing x items is 35x+300 and the revenue for selling x items is 75x−0.5x2 .
The profit that the company makes is how much i
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-> SOLUTION: For a certain company, the cost for producing x items is 35x+300 and the revenue for selling x items is 75x−0.5x2 .
The profit that the company makes is how much i
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Question 1194341: For a certain company, the cost for producing x items is 35x+300 and the revenue for selling x items is 75x−0.5x2 .
The profit that the company makes is how much it takes in (revenue) minus how much it spends (cost). In economic models, one typically assumes that a company wants to maximize its profit, or at least wants to make a profit!
Part a: Set up an expression for the profit from producing and selling x items. We assume that the company sells all of the items that it produces. (Hint: it is a quadratic polynomial.)
Part b: Find two values of x that will create a profit of $300 .
The field below accepts a list of numbers or formulas separated by semicolons (e.g. 2;4;6 or x+1;x−1 ). The order of the list does not matter. To enter a−−√ , type sqrt(a). Answer by math_tutor2020(3817) (Show Source):
Set the profit expression equal to 300 and solve for x.
-0.5x^2 + 40x - 300 = 300
-0.5x^2 + 40x - 300 - 300 = 0
-0.5x^2 + 40x - 600 = 0
Factoring may be possible, but I prefer to use the quadratic formula.
Plug in a = -0.5, b = 40, c = -600
or
or
or
Another approach you can take is to graph these two equations
y = -0.5x^2 + 40x - 300
y = 300
using a tool like Desmos https://www.desmos.com/calculator/jfde1smu5r
In the link above, the graph is shown. The points of intersection will give us the x solutions. Ignore the y coordinates of each point.
You can click on the intersection point to have the coordinates show up.
The intersections of (20,300) and (60,300) lead to the solutions found earlier (x = 20 and x = 60).
Interpretation: If the company sells either 20 items or 60 items, then they will have a profit of $300.