 This lesson is on optimization and modeling. And this time we are going to model word problems. So let's review our tips. The first one is to make sure you read the problem very carefully several times. Not once, not twice, but many times until you truly understand what it is asking for. The second one is to make sure you know what quantity or function is to be optimized. That is very important because in several of the problems you will have more than one function to deal with. The third thing is, if possible, make sketches. A picture is worth a thousand words as you well know. So make sketches on how the elements are related and how the variables change. The next item is to obtain a formula with one independent variable. In related rates you are able to use two independent variables at different times, but in optimization problems we only need one. So everything has to be transformed into one independent variable. The next one says find the critical points and evaluate these at the end points. So make sure you look for your critical points, see the end points, make sure your optimization piece is not at an end point. The next item is if you use calculus, make sure you use a test. And the test is either the F prime test or the F double prime test. And last but not least, just make sure that you answer the question that the problem is asking. Just because you have found X doesn't mean that you have finalized your answer. Let's start our first problem. The first problem is a basic box problem. And the problem reads you are given a 24-inch by 30-inch piece of cardboard. You want to make a box with an open top, which has the maximum volume for the amount of material you are given. What will the dimensions of the squares be that you cut off? So this would be 24 inches, this would be 36 inches, and we are cutting off little squares. So in essence, the bottom of our box will not be 24 by 36, but one side will be 36 minus the amount that you are cutting off. So let's say we call the amount on each side X so you are going to cut off 2X worth for each side. On that side, instead of the bottom part being 24 inches, it's going to be 24 minus 2X. Again, each side of the square being X and you are cutting off X each time. So our formula is volume equals length which is 36 minus 2X times width which is 24 minus 2X times the height which we created to be X. Now because the numbers are so large here, I am not going to continue this one on with calculus, I'm just going to use my calculator to solve this. And many times in pre-calculus classes, you have done that. So let's go to our Y equal to and put in our formula 36 minus 2X times 24 minus 2X times X. And we need a special window for this. So let's go from 0 to 10 and 0 to 2000 and see if that works. Let's do this one by hundreds. And we see we have part of a graph. This looks something like a parabola, but it isn't because you know that volume would be a cubic function and this is just part of the cubic. And what we are looking for is the maximum on this. So we just go to number 4 under calc to the left, to the right. Make sure we give it enough space in between and we find out that the maximum that we will be cutting off is 4.708 inches. So let's go back to our problem and write in X is equal to 4.708 inches. And let's look back. What will the dimensions of the squares be that you will cut off? It will be 4.708 inches by 4.708 inches. Many of you hopefully have done this problem in your pre-calculus class where you've actually cut out the squares from a piece of paper and made the box. You may not study this problem very carefully. Let's go on to another problem, the cone problem. This is another problem that could have been done in a pre-calculus class, but maybe not. Again, it is an optimization problem and can be done with calculus or on your graphing calculator. So this problem reads you are given a circle with a radius of 10 centimeters. What is the volume of the largest cone you can make from this circle? So we have the circle. The radius is 10 centimeters. And we are going to make that into a nice cone. And you begin to understand that this radius is sitting right in here. It is not the radius of the cone. It is actually the side of the cone. So your 10 centimeters sits right there. So your cone has its own radius, which we'll call r, and its own height, which we'll call h. The volume of a cone is one-third pi r squared h. And if you remember in the tips, it says we have to make one independent variable on this right-hand side because we are maximizing volume. We need either r's or h's on the right-hand side. We need to transform our r squared and h to either r or h by using what we know from our cone. So let's just take the piece that looks like a triangle and we know this is 10 and the h is just plain old h. But we can make the r into the square root of 10 squared minus h squared. So that now we can substitute for h in our original equation and get one-third pi. Well, r squared will be the square root squared, so that will be just 100 minus h squared times h. And that equals one-third pi times 100 h minus h cubed. This time I'm going to use calculus on this. So we are now going to take the derivative. So v prime is equal to one-third pi 100 minus 3h squared. And of course, maximum and minimums. The tangent line is horizontal, which means that equals 0. And we will solve for h and we get h is equal to plus or minus 10 over the square root of 3. As you well know, the minus cannot be used because we do not have negative lengths. So we will go with the positive one. Since we did this by calculus, we have to test it. The best test for this particular problem happens to be the double prime test because this derivative is very easy to take. So I will do the v double prime test for you. v double prime is going to be one-third pi and the 100 becomes 0 with the derivative and then the negative 3h squared becomes negative 6h. And when h is positive, v double prime will be negative. Therefore, we have the function which is concave down which creates a maximum. So we do have the largest volume. We don't have to worry about end points on this because the number we have is less than 10 as you can well see. So we want the volume of this largest cone. So now we have to fill in the numbers for volume. So we know volume equals one-third pi r squared h. So volume equals one-third pi r squared is 100 minus 100 over 3 and times h which we already figured out to be 10 over the square root of 3. And computing that we get 2,000 pi over 9 square roots of 3 centimeters cubed or 403.067 cubic centimeters. This is a basic cone problem. Study this one before you do your homework. We have many of these kinds of optimization problems and calculus. This one reads, a landscape architect plans to enclose a 3,000 square foot rectangular region in a botanical garden. She will use shrubs costing $25 per foot along three sides and fencing costing $10 per foot along the fourth side to find the minimum total cost. Always make a picture. We have this rectangular garden. We know what the area is it's 3,000 square feet. Says she will use shrubs costing $25 per square foot along three sides. So let's call these two sides x these two sides y. So on three sides we need $25 per foot and on the third side the fence will cost $25 per foot. And we want to find the minimum total cost. So our cost function is $25 times 2x plus y plus 10y. And again, we have two independent variables sitting there. So we really want to put it into one independent variable. So we will substitute we know that the area is 3,000 so we know area is 3,000 square feet and that's equal to xy. So we can solve for y and have that equal to 3,000 over x. And substitute for y in our cost function. So the cost is now equal to and we can combine all of this. So we'll have 50x plus 35y or 35 times 3,000 over x. If we do the derivative on this we will get cost prime, which is c prime is equal to 50 minus multiply out we get 105000 over x. We get 50x squared is equal to 105,000 x squared is equal to 2100 or x is equal to 10 square roots of 21. And computing that on our calculator we get 45.826 and we are in feet. Remember this is divide. So let's go back to see if we need to test this and of course we do. So let's go to c double prime test again. Second derivative test for concavity and on the 50 we get 0 and that's equal to positive 2 times 105000 over x cubed. If we put a positive number in which again we have here remember the negatives are hardly ever used on these kind of problems because we are using dimensions. So this will be positive which means concave up which means we have a minimum. So now that we have a minimum we know we have that minimum we know what the x is. Let's find out what we have to determine. It says find the minimum total of the minimum total cost will be 50 times 10 square roots of 21 plus 105000 over 2100 or 4582 dollars and 58 cents. Another problem completed in its entirety again when you do calculus test just remember you need to test when you use your calculus. When you use your graphing calculator you can just say that the minimum is at a certain point. Let's do another box problem. This one is quite different from the last one. This one reads 750 centimeter square is the material available for making a box with a square bottom and an open top. What is the largest volume that this box can be? Again make a picture. So here's our box and because it has a square bottom we'll make this side x square bottom and then the height will be something different and of course we'll call that y. So the volume of our box will be x squared y and we are looking for the largest volume so we're going to hold this equation until we find a substitution for either the x squared or the y. And of course that comes from the 750. 750 is the material which means this is the surface area of our box. So we say 750 is equal to the bottom of the box its surface area is x squared and the size of which there are four is x y. Solving for y we get 750 minus x squared over 4x. Putting this back into our volume formula we get volume is equal to x squared times 750 minus x squared all over 4x. Simplifying all of this we get the 750 over 4 can be reduced to 375 over 2x minus and then we multiply the x squared times the x squared divided by x we get x cubed over 4. So now that we have our formula in one variable we do have the choice of going to the calculator or doing it with calculus. I'm going to do it with calculus. So let's do our v prime. So v prime is equal to 375 over 2 minus 3x squared over 4. Set that equal to 0. Solve for x and we get x is equal to again the plus or minus but throw out the minus 5 square roots of 10. Since we did a calculus style we do have to test. So let's use again our second derivative test on this and that is v double prime is equal to negative 6x over 4. Any positive number we put into x would make this a negative so that's negative which means it's concave down which means we have a maximum. So we know we have a maximum. We have our x coordinate. Let's see what the question asks for. What is the largest volume that this box can be? Again your volume is equal to x squared y. So if we fill in for x squared we get 25 times 10. If we fill in for the y we get 500 over 20 square roots of 10. Simplifying all that we should get 6,250 over the square root of 10 and if we stick that in our calculators and get a nice number for we get 1976 .424 centimeters cubed for our volume. So a second kind of box problem a little different from the first one. There are a lot of these problems because of the fact that optimization tends to be one of the more difficult applications of the derivative and that's why I am presenting a lot of these problems to you. So let's look at a cylinder in a sphere. The problem reads determine the dimensions of the largest cylinder that can be inscribed in a sphere whose radius is 10 centimeters. Again we want a picture. So we have a cylinder in a sphere. So we create the sphere and the website will even help you set this up. Now what's nice about a cylinder the volume is equal to pi r squared h and if we look at this h and find the midpoint where the center of the sphere is if we pull this out to the edge we will have a radius and that radius can be used in solving our problem in this case we said the radius was 10 centimeters. So with that in mind with our volume we need pi r squared in h and if you look at this we have two heights here we can call this a half a height or we can call this h and this h and when we put it into our formula we will have the volume equaling pi r squared h and that might be the way to make this a little bit easier for us. Again we need to make a substitution and obviously I'm heading towards the r squared substitution so if I make that radius there that will become because of our Pythagorean there I'm the square root of 100 minus h squared so when I substitute it into my volume formula I'll have volume is equal to pi times 100 minus h squared times 2h and multiplying all of that out we have 2pi which we keep on the outside because we will not use it too much when we are setting this equal to 0 eventually times 100h minus hq again take your v prime and you get 2pi times 100 minus 3h squared equal to 0 h now becomes plus or minus 10 over the square root of 3 again we don't use the minus this time I'm just going to use the v prime test we haven't used that yet and it's available anytime you want to use it and with the v prime or the first derivative test what we do is put our h on a number line 10 over the square root of 3 and decide what happens before that number and after that number this is your v prime test so once we decide we'll know whether the function is increasing or decreasing or just increasing or just decreasing and if we put in numbers smaller than 10 over the square root of 3 we find out that this is positive and if we put in numbers larger than 10 over the square root of 3 we get a negative value which means increasing decreasing we can write the words increasing we can write the word decreasing which means when you're going from increasing to decreasing you have a maximum so we have checked everything out our 10 over the square root of 3 is certainly less than the 10 of the hypotenuse there so what do we do to answer the question determine the dimensions so we already have 1, we have the height but we need to double it in order to find the real height so the real height is 20 over the square root of 3 and then the radius is equal to the square root of 100 minus 100 over 3 times 10 times the square root of 2 over 3 and those are our two dimensions so we have 10 times the square root of 2 thirds centimeters by 20 over the square root of 3 centimeters and that's our final answer on this problem let's go to another problem closest points on a parabola I have a very definite style in doing these you may not see it in all the books the same way I do it but I think my way is quite easy what we are looking for in this particular problem is find the points on the parabola y is equal to x squared minus x plus 5 that are closest to the point negative 7, 5 let's just graph this out before we actually do it so let's look at our function on our calculator put in x squared minus x plus 5 do a zoom 6 on it and it looks something like that so we'll just put that in and then let's plot the point negative 7, 5 so it would be about in here and so we want some point that sits approximately over there and our distance between two points formula is d squared is equal to x sub 2 minus x sub 1 quantity squared plus y sub 2 minus y sub 1 quantity squared I'm going to leave this in the d squared form normally we would have d by itself but this one I'm going to leave in the d squared form so this is equal to we have our negative 7 is our x minus the point on the parabola quantity squared plus 5 minus some point that is y on the parabola and we know we can just substitute for that x squared minus x plus 5 and we want to square that if we take the derivative of this we take 2d d dx and just leave that as it is and the next piece will be 2 times negative 7 minus x to the first power times negative 1 plus 2 times 5 I'm just going to multiply this through negative x squared plus x minus 5 to the first power and that multiplies by negative 2x plus 1 set all that equal to 0 do some algebra on it and you should get x to equal negative 1 again we can do a test on this one once we find our x and this time we are going to do the second derivative test you had 2x cubed minus 3x squared plus 2x plus 7 equals 0 if we do the d double prime test it doesn't matter what goes on that side that's going to equal to 6x squared minus 6x plus 2 and if we put a negative 1 in for x this would be 6 plus 6 plus 2 which is positive this time we actually had to substitute our number in to find out whether it was positive or negative which means we are concave up which means we also have a minimum so we have all the bits and pieces of this it's been tested now we just have to answer the question find the points on the parabola so we have our x coordinate to be negative 1 we just need to determine our y coordinate and if y is equal to negative 1 squared minus negative 1 plus 5 and that all equals 7 so the closest point is negative 1 7 the last problem carrying a pole around a corner a pole that cannot be tilted is carried horizontally around a corner the pole leaves a hole which is 3 feet wide and enters one that is 4 feet wide what is the maximum length of the pole now this kind of problem is presented in your book but is presented with trigonometry I want to present it a little bit differently so just make our diagram so we have 3 feet here and 4 feet here and the pole has to be horizontal as it goes around the corner so it should look like that so if we look at this we can make triangles and we can make similar triangles and break our pole up into two pieces called this piece L1 and this piece L2 and let's have the height of this triangle be y and this length be x so what we are really looking for is L1 plus L2 and like I said we can make similar triangles on this so let me just make those triangles and they're both right triangles and in this one this will be 4 this would be x this would be L2 this would be y this would be 3 this would be L1 so now we could say L2 is to 4 as L1 is to y but y is equal to the square root of L1 squared minus 9 so we can now say that L2 is equal to 4 times L1 over the square root of L1 squared minus 9 so now we can put this together and say the total length is equal to L1 plus L2 which is 4 L1's over the square root of L1 squared minus 9 later and get a final answer and you'll find that L1 will be equal to 4.461 feet and total L will be equal to 9.866 feet and if you put this in your calculator use the x's in max window to be 0 to 20 and the y's min and max 0 to 20 you should be able to find your answer through that this concludes modeling optimization problems