Calculus and Optimization Study Guide
Calculus and Optimization
Knowing the minimum and maximum points of a function is useful for graphing and even more useful in real-life situations. Businesses want to maximize their profits, builders want to minimize their costs, drivers want to minimize distances, and people want to get the most for their money. If we can represent a situation with a function, then the derivative will help find the optimal point.
If the derivative is zero or undefined at exactly one point, then this is very likely to be the optimal point. The first derivative test states that if the function increases before that point and decreases afterward, it is maximal (see Figure 15.1). Similarly, if the function decreases before the point and increases afterward, then the point is an absolute minimum.
The second derivative test states that if the second derivative is positive, then the function curves up, so a point of slope zero must be a minimum (see Figure 15.2). Similarly, if the second derivative is negative, the point of slope zero must be the highest point on the graph. Remember that we are assuming that only one point has slope zero or an undefined derivative.
If there are several points of slope zero and the function has a closed interval for a domain, then plug all the critical points (points of slope zero, points of undefined derivative, and the two endpoints of the interval) into the original function. The point with the highest y-value will be the absolute maximum, and the one with the smallest y-value will be the absolute minimum.
A manager calculates that when x employees are working at the same time, the store makes a profit of P(x) = 15x2 – 48x – x3dollars each hour. If there are ten employees and at least one must be working at any given time, how many employees should be scheduled to maximize profit?
This is an instance of a function on a closed interval because 1 ≤ x ≤ 10 limits the options for x. The derivative of the profit function is P'(x) = 30x – 48 – 3x2 which factors into P'(x) = –3(x2 – 10x + 16) = –3(x – 2)(x – 8). Thus, the derivative is zero at x = 2 and at x = 8.
Because more than two points have a slope of zero, we cannot use the first or second derivative tests. Instead, we evaluate each of our critical points. These are the points of slope zero, x = 2 and x = 8, plus the endpoints of our interval x = 1 and x = 10. These are evaluated as follows: P(1) = –34, P(2) = –44, P(8) = 64, and P(10) = 20. If the manager wants to maximize the store profit, eight employees should be scheduled at the time, because this will result in a maximal profit of $64 each hour.
A coffee shop owner calculates that if she sells cookies at $p each, she will sell cookies each day. If it costs P her 20¢ to make each cookie, what price P will give her the greatest profit?
The function for profit is Profit = Revenue – Costs. If she charges $p per cookie, then she'll make and sell cookies each day. Thus, her revenue will be and her costs will be Therefore, her profit function is We limit this to p > 0.20 because the only optimal situation would be when the cookies were sold for more than it cost to make them.
The derivative is , which is zero when and therefore 80 p2 = 200 p3, so either p = 0 or else p = .Because p = 0 is not in the domain, the only place where the derivative is zero is at p = 40¢
Using the first derivative test, we see that Profit'(0.30) = 740 and Profit'(0.50) = –160, therefore our sign diagram for Profit' is as shown in Figure 15.3. So the absolute maximal profit occurs when the cookies are sold at 40¢.
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