Solution: Maximum and Minimum #2
Solution: Maximum and Minimum #2
The cost function for producing [latex]x[/latex] units of product is given by:
[latex]\Large{C(x) = 100 - 5x + \frac{x^2}{10}}[/latex]
Find the number of units produced that minimizes the cost.
Step 1:
To find the number of units produced (the x value) that minimizes the cost, we will need to find the minimum value of the given function.
To do this, we will find the derivative, and set it to zero. From our knowledge of derivatives, we can easily find the derivative to be:
[latex]\large{C'(x) = \frac{x}{5} – 5 }[/latex]
Step 2:
Now, setting it equal to zero, we have:
[latex]\frac{x}{5} – 5 = 0[/latex]
and solving for x:
[latex]\large{x = 25}[/latex]
This tells us that the minimum value of the function is at x = 25. Thus, 25 units of product are needed to be produced to minimize the cost.
Step 3:
We can graph the function, as shown below, to confirm this:
Solution Complete!
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Solution: Maximum and Minimum #2
The cost function for producing [latex]x[/latex] units of product is given by:
[latex]\Large{C(x) = 100 - 5x + \frac{x^2}{10}}[/latex]
Find the number of units produced that minimizes the cost.
Step 1:
To find the number of units produced (the x value) that minimizes the cost, we will need to find the minimum value of the given function.
To do this, we will find the derivative, and set it to zero. From our knowledge of derivatives, we can easily find the derivative to be:
[latex]\large{C'(x) = \frac{x}{5} – 5 }[/latex]
Step 2:
Now, setting it equal to zero, we have:
[latex]\frac{x}{5} – 5 = 0[/latex]
and solving for x:
[latex]\large{x = 25}[/latex]
This tells us that the minimum value of the function is at x = 25. Thus, 25 units of product are needed to be produced to minimize the cost.
Step 3:
We can graph the function, as shown below, to confirm this:
Solution Complete!
Send us a review!
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