Calculus > How to find minimum profit with calculus

One of the many practical applications of calculus comes in the form of identifying the **maximum** or **minimum** values of a function. Finding that minimum value is how to find minimum profit.

If you’ve spent any time at all in the world of mathematics, then you’ve probably seen your fair share of graphs with attached functions. Many graphs have certain points that we can identify as ‘maxima’ and ‘minima’, which are clearly identified as being the highest or lowest points on a graph. It is quite common to receive a function without an attached graph, so it can be useful to know the method behind getting these values.

## How to Find Minimum Profit with Calculus: Steps

Sample Problem:Identify the minimum profits for company x, whose profit function is f(t) = 100t^{2} – 50t + 9, where ‘f(t)’ is the money gained and ‘t’ is time.

Step 1: Differentiate your function. The reasoning behind differentiating your function is simple: While the function itself represents the total money gained, the differentiated function provides us with the rate at which money is acquired. This will be useful in the next step. Here, f(t) = 100t^{2} – 50t + 9 is differentiated to become f ‘(t) = 200t – 50.

Step 2: Set the equation equal to zero and **solve for t**. In this case, 0 = 200t – 50 becomes 50 = 200t, and t is easily solved to be t = 1/4. This value means that there is either a maxima or a minima at t = 1/4.

Step 3: Test the surrounding values of t (in your original equation) to decide whether your value is a **maxima or a minima.** Here, we plug in our value for ‘t’ in the original equation and identify this point of either maxima or minima. For t = 1/4, this means f(t) = 100(1/4)^{2} – 50(1/4) + 9 = 2.75. Now we pick two very close points to the location of our extrema (t = 1/4). It is important to pick one value greater than and one less than your extrema. Typically, it is wise to pick quick and easy values for this part of the procedure. For instance, 0 and 1 are great choices, not only because they are very close, but also because they will allow you to do the computation in your head. At t = 0, the equation becomes f(t) = 100(0)^{2} – 50(0) + 9 = 9. At t = 1, the equation becomes 100(1)^{2} – 50(1) + 9 = 59.

Step 4: **Compare the results.** If we were to plot our three data points, it would look something like this: (0, 9), (1/4, 2.75), (2,59). You should be able to quickly draw a rough sketch of what this looks like – what you’ll find is that there is a minimum at 1/4. This fact is supported by the fact that the data points immediately to the left and the right of this value are both higher. If they were lower, the point would be a maxima, and if one were higher and the other lower, it would just be a point where the slope of the function is zero.

*That’s how to find minimum profit!*

Warning: Finding the minima of a function is fairly straightforward – but beware, in more complex equations, it can be quite difficult to obtain all of the values for ‘t’ where the function equals zero. In the event that there are multiple values for ‘t’, simple trial and error will lead the way to your minima or maxima.

Note:Step 2 at first seems a little strange, but remember that the derivative of a function represents the rate of the increase or decrease of the original function. If the function representing this rate is equal to zero, that means the actual function is not increasing or decreasing at that specific point. It is likely that at the point where the slope is zero, there will either be maxima or minima to identify.

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