To clarify, imagine a that opens downward. For example, look at the following function and its graph: x y f' x f'' x -2 -1 4 -2 0 3 0 -2 2 -1 -4 -2 Note that a negative second derivative means that the first derivative is always decreasing for a given positive change in x, i. First, let's find the cost to produce 1500 tires. This function is typically called either the demand function or the price function. Therefore, the slope at the top or turning point of this concave function must be zero. So c bar prime, the derivative is going to be 0.
We can use that in conjunction with the first derivative at increasing points of x as you travel left to right on the graph to determine identifying characteristics of functions. I am very confused with this question. Let's take a more in depth look at the cost function and see how it works. If so, are you given an interval of time or something over which you are finding the average? Suppose the function is a higher order polynomial, one that takes the shape of a curve with 2 or more turning points. Obviously, if the average ends up being 19, and the first unit cost 10, then the cost of producing a unit must be changing as we produce different units. Or are you finding the average cost of x units, in terms of the number of units? So wherever there is a critical point, it has an absolute minimum. Hopefully, you remember that the minimum, or maximum, of any function can be found by taking its derivative and setting that equal to zero to find the critical points.
This is the necessary, first-order condition. The word unconstrained refers to the fact that we placed no constraints on the functional relationships we were optimizing. One last note: the title of this section was unconstrained optimization. Now where is this undefined? So the feasible domain is going to have to be x greater than zero. Therefore, we would expect the underlying function to be one where the first derivative is zero at the turning point, with a positive second derivative in the neighborhood of the turning point, indicating an increasing slope.
This is also known as evaluating the second derivative at the critical point s , and provides the sufficient, second-order condition. Now let's see how you would actually use the function. Use the following characteristics to determine whether the function evaluated at the critical point or points is a relative maximum or minimum: Relative Maximum Relative Minimum You will probably always practice on functions where the maximum or minimum does exist, but keep in mind that you will be doing public policy in the real world. The base will cost 9 dollars a square foot and the sides will cost 11 dollars a square foot. These three equations now give us a considerable amount of information regarding the cost process, in a very clear format. Finally, you can find how many items will minimize or maximize the total production cost. Well for x greater than zero, this quantity is always positive and therefore the whole second derivative is always positive.
It would be easy to imagine a function where part was below the horizontal tangent line, turned again, and came back up past the line. Note how much care is being taken to limit the discussion of concavity to the part of the function near the point being considered. Now average cost, this quantity c bar, is defined as the total cost c x divided by the number of units produced, x. In order to fully appreciate this result, let's consider the opposite--a convex function, i. Now, consider the point at the very top of the parabola.
Of course, this is not realistic, and as our models become more realistic in the multivariate section, we will add to our optimization problems. The slope is defined as the rate of change in the Y variable total cost, in this case for a given change in the X variable Q, or units of the good. Example 3: Find the optimum points of the profit function and determine what level of production Q will maximize profit. So this is minus 1000x to the -2. There is no point in doing constrained optimization in univariate processes because it is always easier to embed the constraint within one of the equations and use the same process as outlined in this section.
The ability to use calculus to find minima and maxima is very useful in many areas of study. We see there are two imaginary solutions and one real solution. For example, economic goals could include maximizing profit, minimizing cost, or maximizing utility, among others. For example, calculate the marginal cost of producing the 100th unit of this good. The second derivative being always positive means that it does this kind of a thing. As we often want to minimize average cost in business, it becomes important to business problem solving to recognize that: Example Problem: Given the cost function: a Find the average cost and marginal cost functions. Note that we were able to prove average cost is minimized when Q is 12, without having to actually determine the average cost.
There are some very real applications to calculus that are in the business world and at some level that is the point of this section. Do not forget that there are all sorts of maintenance costs and that the more tenants renting apartments the more the maintenance costs will be. Here are the revenue and profit functions. This kind of analysis can help them determine just what they need to do to move towards that goal whether it be raising rent or finding a way to reduce maintenance costs. In this section we took a brief look at some of the ideas in the business world that involve calculus. You can determine how much it will cost to produce a certain number of items. Therefore, taking the first derivative, or calculating the formula for the slope can determine the marginal cost for a particular good.
Additional Uses of the Cost Functions Besides the total cost, you can use the cost function to find the average cost and marginal cost of production. To find the marginal cost of producing the 1500th tire, we can take the total cost of producing 1500 tires and subtract from that the total cost of producing 1499 tires. If the latter, then try using some actual numbers, and then noting how you operate with them. Example 2: Given the following total cost function, determine the level of production that minimizes the average cost, and the level that minimizes the marginal cost: Solution 2: Convert the total cost function into an average cost function by dividing by Q: Now, to minimize the average cost function, follow the steps listed above. A function, at a given point, is defined as concave if the function lies below the tangent line near that point. So this is the second derivative.
In other words, for an instant, the slope must be zero. You can also determine the average cost to produce all of those items and the marginal cost of producing a specific item. In oil pipeline construction, the cost of pipe to go underwater is 60% more than the cost of pipe used in dry-land situations. The Cost Function: An Introduction How do companies determine the price that they charge to sell certain goods? Although it might seem random, companies frequently use a cost function to determine how many units of an item they should produce and what price they should sell it for. Moving left to right, note that the slope is negative, goes through zero at the turning point, then becomes positive. Now, to minimize marginal cost.