Tuesday, November 28, 2006

4.7: Optimization Problems!

Hi, everybody. I hope the second day back from cozy break hasn’t been too jarring. Now, it’s that happy time again…time for me to do my blog, and this time it’s on Optimization Problems (which are quite similar to Related Rates, my previous blog).

Optimization are all about maximizing and minimizing something. To untangle all the information given to us in those tedious word problems, we can use a series of steps (Related Rates, anyone?):

1. Read the question carefully. Understand the information you are given and, most importantly, understand what is being asked.

2. Draw a diagram. Label everything. This is not absolutely necessary, but it’s usually helpful.

3. Set up a maximum or minimum equation. This step is kind of like identifying the static equation in related rates. This is the equation that you will modify (Step 5) and differentiate (Step 6).

4. Identify the constraints presented in the problem.

5. Use the constraints to rewrite the equation in terms of a single variable.

6. Find the derivative of the equation. Use this to identify critical points. Remember, critical points are wherever the derivative is equal to zero OR where the derivative is undefined.

7. Test the critical points you found, and the endpoints (these will usually have to do with the constraints), to determine which gives the absolute maximum or minimum.

And, of course, don’t forget to answer the question specifically being asked.

All right, since these are application problems, examples are always extremely helpful.

Find the point on the parabola y2 = 2x that is closest to the point (1,4).

We’re looking for a point (x,y). Since we’re looking for closest distance, we can set up a minimum equation using the distance formula:

d = rad[(x -1)2 + (y – 4)2]

This is the distance between the given point (1,4) and the unknown point (x,y).

We are given the equation for the parabola (this is a constraint). We can use this information to solve for x and get rid of the x variables in the equation so we can have an equation in terms of a single variable. (You could also solve for y and substitute that into the minimum equation). Solving for x:

y2 = 2x
x = (y2) / 2

We can substitute this value of x into the minimum equation:

d = rad[((1/2)y2 – 1)2 + (y – 4)2]

To make this equation a little more manageable, we can square it to get rid of the radical:

d2 = [(1/2)y2 – 1]2 + (y – 4)2 = f(y)

Now we can differentiate:

f’(y) = 2[(1/2)y2 – 1](y) + 2(y – 4)(1) = y3 – 8

Don’t forget about the Chain Rule!!

Now, we have to find our critical points. y3 – 8 is not undefined anywhere. We can look for where it equals 0, though.

y3 – 8 = 0
y3 = 8
y = 2

Now we can use the first derivative test to determine whether y = 2 is an absolute maximum or minimum. You can graph the derivative, or simply think it through. When y is negative, y3 – 8 is less than 2. When y is positive, y3 – 8 is greater than 2. By the first derivative test for absolute extreme values, this means that 2 must be the absolute minimum of f(y).

We have our y value for the point closest to (1,4). To find our x value, we plug in our y value into the equation we solved for x.

x = (22) / 2 = 2

So (answering the original question): the point on y2 = 2x closest to (1,4) is (2,2).

Find the area of the largest rectangle that can be inscribed in a semicircle of radius r.

We can draw a picture to make the semicircle the upper half of a circle x2 + y2 = r2 with its center being the origin. The rectangle inside would have a width of y on either side and a length of 2x on either side. Thus, the area is:

A = 2xy

Now we want only one variable. We can solve for y using the equation of a circle:
y2 = r2 – x2
y = rad(r2 – x2)

Substituting this back into our area equation:

A = 2x[rad(r2 – x2)] domain: [0, r]

Now we differentiate the area equation:

A’ = 2 rad(r2 – x2) – (2x)(1/2)(r2 – x2)^(-1/2)(2x)
A’ = 2 rad(r2 – x2) – (2x2)(r2 – x2)^(-1/2) = 2 rad(r2 – x2) – (2x2) / [rad(r2 – x2)]
A’ = [2(r2 – x2) – 2x2] / [rad(r2 – x2)]
A’ = [2(r2 – 2x2)] / [rad(r2 – x2)]

To find the critical points, we look at where A’ equals 0:
2(r2 – 2x2) = 0
2r2 – 4x2 = 0
4x2 = 2r2
x = r / (rad2)

Plug this value of x into the original area equation (remembering that we also solved for y earlier, so we have something to plug in for x and y):

A = 2[r / (rad2)][rad(r2 – x2)]
A = 2[r / (rad2)][rad(r2 – r2/2)]
A = r2

Before we assume this is our maximum value, we observe the behavior of the endpoints:

A(0) = 2(0)[rad(r2 – 02)]
A(0) = 0
A(r) = 2(r)[rad(r2 – r2)] = 2r(0) = 0

This confirms that x = r / (rad2) gives us a maximum value of A. Now, to answer the question: the area of the largest inscribed rectangle is r2.

The examples we did in class show the optimization problem procedures using more numbers (like the fence area one), whereas these show optimization problems in the coordinate plane, in a slightly more abstract way. It's useful to see this side of the optimization problems, but it's also good to keep in mind that optimization problems can be applied to real world situations. Our homework definitely gives us a chance to work with some, doesn't it?

Here's a link to a website with a lot of sample problems: good practice.

So, if you thought we had icky math tests...take a look at this:
Ah, yes.

Tessa, you're up next.

Ave atque vale, everyone. ~8)


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