Maximum and Minimum Values Practice Problems
Problem: Maximum and Minimum #1
Determine the maximum and minimum values for the function:
\Large{f(x) = x^3-6x^2+9x}
On the interval: \Large{[-1, 4]}
Step 1:
To find the maximum and minimum values, we must find the critical points. We know that at the critical points, the rate of change, or derivative is zero. Hence, we will find the derivative of the function and set it to zero.
Using the power rule, we can find the derivative to be:
\large{f'(x) = 3x^2-12x+9}
Then, factoring and setting to zero, we have:
\large{ 3(x-)(x-3)= 0}
This tells us that the critical points, and thus, the maximum and minimum values of the function are located at x = 1 and x = 3.
Step 2:
Now, we need to find which one of these is a local maximum and which one is a local minimum.
If we plug in x = 1 into the original function we get 4.
However, if we plug in x = 3, we get 0.
From this, we can determine that the local maximum lies at x = 1, and the local minimum lies at x = 3.
Step 3:
Graphing the function, we can confirm this to be true:
Solution Complete!
Problem: Maximum and Minimum #2
The cost function for producing x units of product is given by:
\Large{C(x) = 100 - 5x + \frac{x^2}{10}}
Find the number of units produced that minimizes the cost.
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