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Aim: How can we apply the first derivative to solve easy, medium and hard level problems related to Optimization?

Aim: How can we apply the first derivative to solve easy, medium and hard level problems related to Optimization?. Donow: Try these compositions of functions. If f(X)=2, f’(3)=-1, g(3)=3, g’(3)=0, Find F’(3). (question borrowed from Mrs. Dela Cruz) F(x)= 2f(x)- g(x)

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Aim: How can we apply the first derivative to solve easy, medium and hard level problems related to Optimization?

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  1. Aim: How can we apply the first derivative to solve easy, medium and hard level problems related to Optimization? Donow: Try these compositions of functions

  2. If f(X)=2, f’(3)=-1, g(3)=3, g’(3)=0, Find F’(3) (question borrowed from Mrs. Dela Cruz) • F(x)= 2f(x)- g(x) Just think here to your self, is there any product rule or division rule involved?

  3. (con). • No right? Good. Let’s simply make the f(x) and g(x) into f’(x) and g’(x) respectively. (dont’ worry about the two!!!) • 2f’(x)-g’(x) = F’(x) Now just plug in: F’(3)= 2f’(3)- g’(3) Based of our charts we see that f’(3)= -1 g’(3)= 0

  4. Continued… • So, its F’(3)= 2(-1)-0= -2 That’s it, see how easy it is… Let’s try another one: F(x)= 1/2f(x)g(x) Ask yourself: Does this question involve the product rule or the division rule Yes!!!!

  5. The Product Rule • Well, don’t panic! There is a simple product rule we can apply: • F’(x)= f’g+g’f. You may say: Here we go again… Well, this derivative is simply the derivative of f times g added to the derivative of g mulitplied with the f value. Don’t get it? Well, lets approach the example:

  6. Problem F’(x)= 1/2[f(x)g’(x)+g(x)f’(x) F’(3) = 1/2[f’(3)+g(3)+g’(3)f(3) F’(3)= ½[2(0)+3(-1)] The algorithm Application of the product rule Placing what needs to be derived, in the equation Plugging in with the values of the original equation Solving: ½ f(x)g(x)…

  7. F’(3)= -3/2 Evaluating the expression. Continuation of the donow prob.

  8. Why the donow problem? • You see what I was trying to show you? Tell me, what was common about both things: The first derivative. For all of them, I was trying to show you that the first derivative plays an integral role in our finding of the derivatives. In our lesson, it will play a bigger role Now on to our lesson…

  9. What is Optimization? Optimization is basically trying to get the most of something while getting rid of the least. It can also be trying to get least amount of something. You see the big oil companies: Exxon Mobil, Chevron Corporation and BP trying to make the most amount of money, but trying to cut their costs in a highly competitive environment.

  10. Optimization functions (con) • Sometimes, we want to get the least volume with the most amount of material, so that we don’t waste any. • As you can see, Optimization has a wide array of applications from big name gas tankers to simple cubic volumes. I’ll help you dissect the problems…

  11. Question 1 Intimidated? Relax, we’ll start with something easy: (borrowed from www.math.ucdavis.edu) Find two nonnegative numbers whose sum is 9 and so that the product of one number and the square of the other number is a maximum.

  12. Approaching the problem You may have difficulty if you don’t break the problem down into simple parts: The problem states: “nonnegative numbers” So I tell myself “Only positive numbers” Next, it says “ sum of 9” We don’t know what the sum of the problem, is so we just have to make them up.

  13. Approaching the problem…con. Let’s say that “x” is the first number, and “y is the second number. So, it follows that x+y=9. Simple enough? Ok, take a deep breath… Now, read the second part of the problem… (the product of one number and the square of the other is a maximum). Seems a little tricky, right?

  14. Continued. We know that one number is “x” and the other number is the “square,” So, this means that the “y” is the y squared. Let P represent the product of the equation: Now, we must attempt to connect both parts, which is probably the toughest part, that I had trouble with…

  15. The connection: • Well, we want to find the product of the equation,so our goal is to make both of the variables in the product the same( You can select either variable you want to put in terms of, but lets use “y”, for simplicity It follows that:

  16. Derivative Time:) • Remember what I had told you before about the first derivative, well now it’s time to apply it… • P’ = f’g+g’f • P' = x (2) ( 9-x)(-1) + (1) ( 9-x)2 • = ( 9-x) [ -2x + ( 9-x) ] • = ( 9-x) [ 9-3x ] • = ( 9-x) (3)[ 3-x ] • = 0 • X=3 or 9.

  17. Plugging back… Remember the plugging back, now we do the same… If we do 9*0 or 0*9, well have 0. So the only ones that work is 3 and 6, which well get 108 as the maximum product.

  18. Example 2: Borrowed from http://www.math.ucdavis.edu Medium Problem: A sheet of cardboard 3 ft. by 4 ft. will be made into a box by cutting equal-sized squares from each corner and folding up the four edges. What will be the dimensions of the box with largest volume ?

  19. Solving the problem Drawing a picture is the safest way to start, draw a picture, and label the sides!

  20. Other steps: • Here is the Rectangular Box without the cover:

  21. Steps: • Observe that the problems that we solve are similar to that of the previous problem. These are: writing the equation, getting the “solved for” equation in terms of the same variable. Finding the first derivative, setting it equal to 0, and then solving for the variable. If we want to check whether the answer is maximum, we plug our answer into the second dx/dy. If the answer is less than 0, were okay, otherwise, it does not exist.

  22. Finally solved… • V =LWH = (4-2x) (3-2x) (x) . • V' = (-2) (3-2x) (x) + (4-2x) (-2) (x) + (4-2x) (3-2x) (1) • = -6x + 4x2 - 8x + 4x2 + 4x2 - 14x + 12 • = 12x2 - 28x + 12 • = 4 ( 3x2 - 7x + 3 ) • = 0 (notice we had to use the product rule three times, since there are 3 “parts • Now, you must use the quadratic quation to solve for the x value. You will find that the answer comes out to x= 0.57 or 1.77. Now, you must plug back in to the original equation. You will find that the largest volume is 3.03 cubic feet at x=0.57.          or          .

  23. Hard problem Is it getting easier? I certainly hope so. Now will come the hardest one of all… Borrowed from http://www.math.ucdavis.edu Consider a rectangle of perimeter 12 inches. Form a cylinder by revolving this rectangle about one of its edges. What dimensions of the rectangle will result in a cylinder of maximum volume ?

  24. Approaching the toughie… • Let’s draw a picture, we know that the rectangle is inside the cylinder, and only half way. So here it is.

  25. Dissecting the problem • We know that the perimeter is the sum of twice the width and twice the length. So, its 12 = 2r + 2h , now, we need to find the equation of the volume of the cylinder. Let p represent pi. V=pr2h. Now, notice that we have “h in both equations. Since we have this, we can make the perimeter equation in terms of r. Now, if we do that, all we have to do, is follow the aforementioned rules of finding the first derivative, setting it equal to 0, and then plugging in. We should then have the maximum volume.

  26. Solved answer. • V’=p(12r-3r2 ) p(3r)(4-r) 3pr(4-r)=0 r= 0 or 4 Now, you have to guess different values. You'll find that when r=4 and when h=2, there is the highest volume. This volume will be 100 feet cubed.

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