Simple Optimization: Calculus I and II (AB and BC)
Find two numbers whose product is a minimum, given that one number is five less than one-third the other, and justify your answer mathematically.
Difficulty:
Solution:
Simple Optimization: Calculus I and II (AB and BC)
Find two numbers whose product is a minimum, given that one number is five less than one-third the other, and justify your answer mathematically.
Difficulty:
Solution:
We'll call the first unknown number x and the second
. Since you are trying to find the minimum product, the function you want to optimize is the product function (i.e. just multiply them together).
To optimize, first distribute the x through the parentheses (to make differentiation easier), and take the derivative.
The function P has a maximum or minimum whenever P ' = 0, so set it equal to 0 and solve.
Thus, one of the numbers is probably 15/2. To justfify that, you have to draw up a wiggle graph (a number line with the critical numers--in this case only 15/2--marked). Test points higher and lower than 15/2 by plugging them into P '. You'll get this graph:
Because P ' changes from negative to positive at 15/2, the function goes from decreasing to increasing there. Thus, x = 15/2 (or 7.5) is a minimum. However, don't forget to answer the question posed by the problem: What are both of the numbers? One of them is x = 15/2 (or 7.5). The other is
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