# Maxima & Minima Word Problem

Problem: Given the following profit-versus-production function for a certain commodity: $P=200000-x-(\frac{1.1}{1+x})^8$. Where P is the profit and x is the unit of production. Determine the maximum profit.

Solution: Taking its first derivative, $\frac{dP}{dx} = -1-8(\frac{1.1}{1+x}^7) * (\frac{-1.1}{(1+x)^2})$, then equate to $0$, the value of x would be equal to $0.371$. Then substituting it to the original equation would result to $199,999.46$ which is the maximum profit.

Question:

1. How to solve if instead, the problem asked for the minimum profit?
2. In some problems, the minimum is the value of x (example: the 0.371 in the problem above) after differentiating the given equation and equating it to 0. But in some problem the minimum is the value after substituting that x, so in some problem, that 199,999.46 is the minimum instead. So how can I know which is which?

Any help or tip would be appreciated.

• Question 1: For the minimum profit, we would need more information about the range of $x$. It's probably a positive number, right? With $x$ limited to positive numbers, it' possible for $P$ to be zero ($x\approx 200000$) or even negative. Question 2: I would say that the word minimum and maximum are usually related to the value of the function that we're analysing, so it's the value of $P$ and not $x$. – Matti P. Jun 4 '18 at 6:53
• @MattiP. Hmm. There were problems where the solved x is equal to P which is equal to minimum or maximum but I guess there can only be one answer in those kinds of problems. Thank you. – Jayce Jun 4 '18 at 7:37

One way to check which one is Maximum or Minimum is to take the second derivative and check the sign of the second derivative at critical points(https://www.math.hmc.edu/calculus/tutorials/secondderiv/).

By Second Derivative test we can decide function is maximum or minimum at point x. http://mathworld.wolfram.com/SecondDerivativeTest.html http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-maxmin-2009-1.pdf