Question

what's maximization and minimization in coordinate geometry? please explain with an example.... 30 pts

Answer 1

Minimization and maximization refresher

 

The fundamental idea which makes calculus useful in understanding problems of maximizing and minimizing things is that at a peak of the graph of a function, or at the bottom of a trough, the tangent is horizontal. That is, the derivative f′(xo)f′(xo) is 00 at points xoxo at which f(xo)f(xo) is a maximum or a minimum.

Well, a little sharpening of this is necessary: sometimes for either natural or artificial reasons the variable xx is restricted to some interval [a,b][a,b]. In that case, we can say that the maximum and minimum values of ff on the interval [a,b][a,b] occur among the list of critical points and endpoints of the interval.

And, if there are points where ff is not differentiable, or is discontinuous, then these have to be added in, too. But let's stick with the basic idea, and just ignore some of these complications.

Let's describe a systematic procedure to find the minimum and maximum values of a function ff on an interval [a,b][a,b].

Solve f′(x)=0f′(x)=0 to find the list of critical points of ff.Exclude any critical points not inside the interval [a,b][a,b].Add to the list the endpoints a,ba,b of the interval (and any points of discontinuity or non-differentiability!)At each point on the list, evaluate the function ff: the biggest number that occurs is the maximum, and the littlest number that occurs is the minimum.

Example

Find the minima and maxima of the function f(x)=x4−8x2+5f(x)=x4−8x2+5 on the interval [−1,3][−1,3]. First, take the derivative and set it equal to zero to solve for critical points: this is

4x3−16x=04x3−16x=0

or, more simply, dividing by 44, it is x3−4x=0x3−4x=0. Luckily, we can see how to factor this: it is

x(x−2)(x+2)x(x−2)(x+2)

So the critical points are −2,0,+2−2,0,+2. Since the interval does not include −2−2, we drop it from our list. And we add to the list the endpoints −1,3−1,3. So the list of numbers to consider as potential spots for minima and maxima are −1,0,2,3−1,0,2,3. Plugging these numbers into the function, we get (in that order) −2,5,−11,14−2,5,−11,14. Therefore, the maximum is 1414, which occurs at x=3x=3, and the minimum is −11−11, which occurs at x=2x=2.

Notice that in the previous example the maximum did not occur at a critical point, but by coincidence did occur at an endpoint.

Answer 2

Answer:

Write an equation for the quantity that is being maximized or minimized (cost, profit, amount, etc.). This is that maximization or minimization equation. ... The coordinates that give the largest or smallest value for this equation (depending on what the problem is looking for) are the solution to the problem.