A manufacturing company finds that the daily cost of producing x items of a product is given by c(x)=210x + 7000. Find x using calculus
Well, let's get down to "business" and find that value of x using calculus!
The cost function c(x) is given as 210x + 7000, where x represents the number of items produced. To find the value of x, we need to minimize the cost. In calculus, we can find the minimum of a function by taking its derivative and setting it equal to zero.
Let's differentiate c(x) with respect to x:
c'(x) = 210
Now, let's set c'(x) equal to zero and solve for x:
210 = 0
Uh-oh! Something seems a bit off here. It looks like we made a mathematical "blunder."
The equation 210 = 0 has no solutions, which means there is no critical point for this cost function. It seems like we have entered a perplexing land of mathematical absurdity.
However, fear not! We can still determine the value of x. Since c(x) represents the daily cost, it will always be positive as the manufacturing process incurs expenses.
So we can't find a minimum, but we can find the value of x by setting c(x) equal to zero and solving for x:
210x + 7000 = 0
This equation can be rearranged to find:
210x = -7000
x = -7000/210
Using a calculator to do the math, we find that x β -33.33.
Well, isn't that amusing? It seems we've landed on a negative value for x, which doesn't make sense in this context. It appears there might be a perplexing error in the problem statement or our calculations.
So, until we can "crack the code" of this challenge, the value of x remains a mystery.
To find x, we can use calculus by finding the derivative of the cost function c(x) with respect to x and set it equal to zero.
c(x) = 210x + 7000
First, let's take the derivative of c(x) with respect to x:
c'(x) = 210
Setting the derivative equal to zero:
210 = 0
Since 210 is a constant and not equal to zero, there is no value of x that will satisfy the equation c'(x)=0.
Hence, there is no critical point in this case.
Therefore, we cannot find x using calculus. However, if you provide additional information or constraints, we may be able to find a specific value for x.
To find the value of x that minimizes the daily cost function, we need to take its derivative and set it equal to zero. Here's how you can do it step by step:
1. Start by finding the derivative of the cost function c(x) with respect to x. The derivative of 210x is simply 210, as the derivative of a constant multiplied by a variable is equal to the constant. The derivative of 7000 with respect to x is zero, as it is a constant term.
2. So, the derivative of c(x) is: c'(x) = 210.
3. Set the derivative equal to zero and solve for x: c'(x) = 0.
210 = 0
Since this equation is impossible to satisfy, we can't find a value of x that would minimize the cost using calculus.
4. However, we can still analyze the behavior of the cost function. As the derivative is positive (210) and does not change sign, we can conclude that the cost function increases linearly with the number of items produced.
Therefore, the cost will continue to increase indefinitely as the number of items produced increases.
x = (c-7000)/210
No calculus needed; just algebra I