Shantilal Shah Engineering College, Bhavnagar

Instrumentation and Control EngineeringSemester 1 (Bridge Course)

Group no. 2 (I.C.)Group no. 2 (I.C.)

Members :Members : Faculty :Faculty :

R.M. Agrawal D.K. Katariya

TopicTopic : :

Concepts of Concepts of Maxima And Maxima And

MinimaMinima

The Least and the GreatestThe Least and the Greatest Many problems that arise in mathematics call

for finding the largest and smallest values that a differentiable function can assume on a particular domain.

There is a strategy for solving these applied problems.

Some HistorySome History Many problems in the seventeenth century that

motivated the development of calculus were maxima and minima problems.

Some of these problems involved finding the maximum range of a cannon (Galileo), the maximum height of a projectile fired from various angles (Galileo) and finding the greatest and least distance of a planet from the sun (Fermat and Descartes).

The Max-Min Theorem for Continuous The Max-Min Theorem for Continuous FunctionsFunctions

If f is a continuous function at every point of a closed interval [a,b], then f takes on a minimum value, m, and a maximum value, M, on [a,b].

In other words, a function that is continuous on a closed interval takes on a maximum and a minimum on that interval.

The Max-Min Theorem, GraphicallyThe Max-Min Theorem, Graphically

Strategy for Max-Min ProblemsStrategy for Max-Min Problems The main problem is setting up the equation:

Draw a picture. Label the parts that are important for the problem. Keep track of what the variables represent.Use a known formula for the quantity to be maximized or minimized.Write an equation. Try to express the quantity that is to be maximized or minimized as a function of a single variable, say y=f (x). This may require some algebra and the use of information from the problem.

Strategy for Max-Min Problems Strategy for Max-Min Problems (continued)(continued)

Find an interval of values for this variable. You need to be mindful of the domain based on restrictions in the problem.Test the critical points and the endpoints. The extreme value of f will be found among the values f takes at the endpoints of the domain and at the points where the derivative is zero or fails to exist.List the values of f at these points. If f has an absolute maximum or minimum on its domain, it will appear on the list. You may have to examine the sign pattern of the derivative or the sign of the second derivative to decide whether a given value represents a max, min or neither.

Using the StrategyUsing the Strategy Suppose that you buy 36 feet of fencing. What are the Suppose that you buy 36 feet of fencing. What are the

dimensions of the rectangular plot of maximum area?dimensions of the rectangular plot of maximum area?Draw a picture:

Length = x

Breadth = y

Using the Strategy (continued)Using the Strategy (continued)Write an equation.

WE KNOW: A= lb = xy. This is the equation to be differentiated, since we are maximizing area. We would like to reduce the number of variables to one.We also know that we have 36 feet of fencing. This tells us the perimeter of the rectangle:

2x + 2y = 362x + 2y = 36NOTE: x cannot be <0 and 2x cannot be >36.

0 0 << x x << 18 18

Using the Strategy (continued)Using the Strategy (continued)Let’s solve the last equation for y. That gives us y=18-x (good old algebra). Substituting this into

A = xy, we get:A = x ( 18-x)= 18 x – x2

Take the derivative of this equation (easy!)

dA-------

dx= 18 – 2x

Using the Strategy (continued)Using the Strategy (continued)Set the derivative to zero and solve to get

x = 9.Take the second derivative of A= 18 x – x2 to get

Since the second derivative is a negative number and a constant, this must be the only maximum.

2

22

d A

dx= −

Using the Strategy Using the Strategy (no, we’re not done yet)(no, we’re not done yet)

At this point, we might think we are done. WRONG!!! The problem is asked for the dimensions of the rectangle.

x=9 gives the length of the rectangle. Since we have y = 18 - x, y=9. The rectangle must be a 9 x 9 square to get the plot of maximum area.REMEMBER--answer the question that is asked!

Some CommentsSome CommentsNote that we only had one point to test. There was only one value when we set the derivative equal to zero and solved the equation and the second derivative test told us that we had found an absolute minimum. We could have checked that value and the endpoints of the domain:

It is clear that the area of the rectangle is maximum at x=9 (y=9). Hence we’ve checked the solution using Linear Programming.

x 0 9 18

A 0 81 0