Calculus Word Problem: Profit

Of course, you can simplify a bit first by dividing both sides by 40: 1620(1/40)\sqrt{x}= 40(1/40)x+ 12200(1/40) But I don't see how you got the equation you have so I can't tell you where you went wrong.In summary, a company needs to determine the number of items it must sell in order to break even. The best price for each item is given by the formula P=1620 sqrt(x)/x, with x representing the number of items sold. The cost to produce each item is $40 and the overhead cost is $12,200. By setting the income equal to the cost, a quadratic
  • #1
AGA19
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Homework Statement



A company sells a certain product and determines that the best price that each item/product must be sold for is given by the formula P=1,620 Square Root X Over X. AKA P=1,620 Square Root X divided by X. X is the # of items sold. Each item cost's $40 to produce4. All items produced are sold and the overhead cost is $12,200. Find the # of items that must be sold for the company to break even.

Homework Equations



12,200=1,620 Square root X/X?

The Attempt at a Solution



Have no idea what I'm doing here. I'm guessing that you plug 12,200 in for P. But then how do you find X? I'm so lost. Please help!
 
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  • #2
1620sqrt(x)/x - formula for how much each item is worth when sold
40x - the cost of every item produced
12200 - overhead

p(x) = 1620sqrt(x)/x - 40x - 12200
1620sqrt(x)/x should be positive as you're making money, and the other two should be negative because you're spending it to start up.

So now find d/dx p(x).

edit: Actually, if it's just to find when the company break even and not an optimize profit, set p(x) = 0 and solve from there.
 
  • #3
So the problem should now look like. 0=1620 square root X/X -40X -12200. So I'd add 12,200 to both sides.

12,200+40 X=1,620 square root X/X?

Sorry Calculus is killing me this semester.
 
  • #4
Can anybody further explain this please?
 
  • #5
You have
[tex]P= \frac{1620\sqrt{x}}{x}[/tex]
but I am puzzled as to why that is not written as
[tex]P= \frac{1620}{\sqrt{x}}[/tex]
or just [itex]P= 1620x^{-1/2}[/itex]

And, indeed, I serious problem with that! It says that the more you raise the price, the more you sell! In order to sell 4 times as much (replace x with 4x) your prices should be twice as much. Normally, raising the price of something discourages people from buying and reduces the amount sold.

Are you sure it wasn't
[tex]P(x)= \frac{1620- \sqrt{x}}{x}[/tex]
or something like that?

Anyway, taking the formula you gave for price, the income from selling x things is just the price times x:
income= [itex]1620\sqrt{x}[/itex]

You say they each cost $40 to make and there are overhead costs of $12200. So making "x" of them costs 40x+ 12200

"Break even" is when income equals cost:
[tex]1620\sqrt{x}= 40x+ 12200[/tex]

There is no "profit" and you don't need calculus to solve that- just square both sides to get a quadratic equation and then solve that equation.
 

1. How is calculus used to solve profit word problems?

Calculus is used to solve profit word problems by analyzing the rate of change, or derivative, of the profit function. This allows for finding the maximum or minimum points, which represent the optimal profit.

2. What information is needed to solve a profit word problem using calculus?

To solve a profit word problem using calculus, you will need the cost and revenue functions, as well as any other relevant information such as fixed costs or the price of the product.

3. Can calculus be used to find the break-even point in a profit word problem?

Yes, calculus can be used to find the break-even point in a profit word problem by setting the profit function equal to zero and solving for the corresponding value of the independent variable.

4. Are there any limitations to using calculus in profit word problems?

One limitation of using calculus in profit word problems is that it assumes a linear relationship between cost and revenue, which may not always be the case in real-world scenarios. Additionally, it may not take into account external factors that could impact profit.

5. How can understanding calculus help in making business decisions?

Understanding calculus can help in making business decisions by providing a quantitative analysis of the profit function, allowing for more informed decisions on pricing, production levels, and other factors that impact profit. It also allows for optimization of profits by finding the maximum or minimum points of the profit function.

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