an employee's monthly productivity in number of units produced is found to be a function of the number of years of service for a certain product a productivity function is given we're looking for the maximum productivity all right whenever we want maximum or minimum that's optimization so we can use calculus what we want to do are we want to find the critical values and also since this specifies an interval we want to also consider the endpoints so we're going to First Look for the critical values and remember that critical values occur when the first derivative is either zero or does not exist so let's go ahead and find that first derivative so our first derivative would be negative 8 T plus 224 and that's a polynomial linear function so it's continuous everywhere in other words there are no places where the derivative does not exist we don't have any candidates for that here but next we want to check where that derivative is equal to zero so let's go ahead and set that equal to zero and solve we get a negative eight t equal to a minus 224 dividing both sides by negative 8 is going to give us a value of 28. so our critical value is 28 we just have the one critical value and then we have the end points 0 and 40. so what we want to do now is check each of these to see which one of those Returns the maximum productivity now we could use the first derivative test to find out whether our critical value is a maximum or a minimum but since we only have one critical value over this interval and since our derivative is continuous over this interval we can also use our second derivative test to find out whether our critical value is a Max or a min so if I go ahead and find our second derivative second derivative here would be a negative eight which is always going to be negative which means concave down and that tells us that the critical value will be a maximum second derivative is negative that's concave down which gives us a critical value that's maximal or you could check it in the first derivative test critical values 28 you wanna choose a test point to the left of 28 choose a test point to the right of 28 and plug those in for example if I wanted to try negative 100 plug that in to the first derivative and I would see I got positive 800 plus 224 which returns a positive value something that's to the right of my critical value if I plug that into my first derivative I'm going to get Negative 800 plus 224 which is going to return a negative value if I'm increasing to the left and decreasing to the right that tells me my critical value is at the top of the hill so to speak which also confirms it as a maximum so I see that this critical value does turn into a relative maximum and then just formally finish it I have two endpoints 0 and 40. and then my one critical value make sure that the critical value is the one that returns the greatest output so if I put 0 back in the original equation I'll get out 150. if I put 28 into the original function I'll return 32.86 and putting 40 into the original function gives an output of 2710 and you can check each of those one by one put these T values into the original function and indeed the highest output goes with the critical value of 28 which we've already determined twice was a maximum so in the year T28 we didn't have a initial or starting year so we just leave it at that in the year of T equals 28 the maximum productivity is 3286. for a dosage of X cubic centimeters cc's of a certain drug the resulting blood pressure B is approximated by the function below find the maximum blood pressure and the dosage at which it occurs again that word maximum tells us optimization so we can go through some of the similar steps that we did earlier we want the critical values critical values occur when the derivative does not exist or when the derivative is equal to zero so go ahead and find that first derivative this is also a polynomial so it's continuous everywhere and especially specifically continuous on the given interval so there are again no places where the derivative does not exist next we'll set that derivative equal to zero and solve to see if we come up with any critical values If I subtract 770x from both sides that was a nice thought but the inclination here is to divide both sides by X but the problem with that step is you don't know if x is zero and if you divide both sides by x algebraically and x turns out to be zero then you've just done something that's undefined so a better approach instead of subtracting the 770x from both sides would be instead to do factoring let's factor out that common factor so I can take out 70x that'll leave me with a 1 here minus 9 x and you can check that by Distributing it back in you certainly can use a calculator to help you factor that out 770 is a factor of 69 30. goes into that nine times all right now since we have it set equal to zero this first part here could turn into zero if x itself was Zero and the second part here could turn into zero if 1 minus 9x is 0. so solving this I get X is equal to positive 1 9. so I have one two critical values coming from setting the first derivative equal to zero and then recall that I also want to check endpoints so you can line those up those critical values up on first derivative test we have zero and then we have 1 9 positive and our negative infinity and now we can choose some test points that are in these respective intervals and test them in our first derivative so go ahead and choose a couple of test points test them in the first derivative so I chose negative one for the first interval that falls between zero and negative Infinity putting negative 1 into my first derivative I got out negative seven thousand seven hundred but what I care about is the fact that it's negative between 0 and 1 9 I used a test point of one-tenth put that into my first derivative and worked it out got an output of 7.7 positive to care about is positive when I tested positive one in the first derivative got an output of negative 6160 which is negative so finishing my first derivative test here I see I have a decreasing followed by an increasing followed by a decreasing so that confirms that 0 is at the bottom of a hill I'm sorry at the bottom of a valley so 0 is a critical a critical value that is also a relative min and one-ninth is at the top of the hill so that's a critical value that is also a relative Max okay then we want to check these values 0 and 1 9. along with our endpoints coincidentally zero we've already got that's one of our endpoints so the other one we want to check is 0.16 we know that zero now we want to check these rather we want to check these into our original equation so one at a time check these input values for X back into the original equation to see what the B of X output values are pause the video and check those in the original equation remember that test points for the first derivative test get checked in the first derivative but when you want to know the actual output from the function you're going to check those critical values and end points back into the original function and I find it easiest to type the function into my calculator in the y equals button and then just skip over to the table and enter the values for x these are the values that I came up with so I'm looking for the maximum output the maximum blood pressure it looks here that the highest output value is 1.58 and that comes when you have one over nine so the critical value of one over nine is the amount of the drug that you want to use to return the maximum blood pressure instantly 1 over 9 is 0.111 approximately as a decimal maximize there's that word again maximize b equals y squared given that X and Y are positive numbers and we also know that X Plus y squared is equal to 9. so this is what we call our objective function that we want to maximize that's our main equation and then this is what's known as a constraint or a side equation that kind of helps us out we also have another constraint here in so far as X and Y must be positive so we can write that as X has to be greater than 0 and Y has to be greater than C so we actually have one two three constraints helping us with our objective function all right so we want to maximize this objective function but in so doing you want to make sure that you only have one input and one output variable and right now we have more we have one output but we have two inputs so we need to clean that up we want to get it to the point where we only have one input variable and one output variable that means either replace x with a y or replace y with an X and how do you do that you do that based upon your side equations you let them help you out so I think it would be easier to get X by itself in my constraint so I'm going to go ahead and set this equal to X by itself by subtracting y squared from both sides and now I can make a substitution b equals x times y squared I can replace the X with what I just discovered it's equal to and that's 9 minus y squared let's replace this x with 9 minus y squared copy down the other y squared and put it together distribute this y squared so I'll have 9y squared minus y to the fourth so that's my main equation or my objective function now notice I haven't done any calculus yet I'm taking any derivatives since I'm asked to maximize I know that's coming I want to get that derivative so that's going to give me 18y minus 4y cubed okay so that's my derivative which I can now use to find my critical values is there anywhere where this derivative does not exist no so I move on to taking that first derivative and setting it equal to zero and now I want to solve looks like I can factor out a variable of Y um and a common factor of two if I take a 2 out of there that leaves me with a 9 take a 2 out leaves me with a 2 I took out one of the Y's so I've got 2 left uh further Factor this that's not quite the difference of two squares although it's close but I know that anything times 0 is 0 so this first item will turn into zero when Y is zero in this second portion will turn into zero well [Music] nine times y squared is equal to nine halves dividing both sides by negative two so now I need the square root of both sides y equals plus or minus 3 over the square root of 2. taking the square root of both sides and that mathematically is an answer but if you were to what's called rationalize the denominator you'd multiply the top and the bottom you'd multiply the top and the bottom of this by square root of 2 to get it into more acceptable or formal format if you will but sometimes this is also acceptable with the radical on the bottom all right so we ended up with one and two two three because plus or minus so we have a total of three critical values and now we want to maximize so we want to see if any of these critical values return a maximum amount we don't have we don't have a domain or an interval so we don't have to check any endpoints we just have to make sure that they meet our constraints speaking of that meeting our constraints both of them must be positive both of them must be positive so that takes out zero it's positive means greater than zero and it also takes out the negative so that leaves us with only one critical value the positive 3 radical two over two or the positive 3 over the square root of 2. since I only have one candidate as a critical value to check I can go ahead and utilize the second derivative and the second derivative coming here from the first derivative is 18 minus 12y squared so go ahead and check this 3 over radical two into the second derivative put 3 over radical 2 in for y see what you get so 18 minus 12 multiplied by 3 over square root of 2 squared that's going to be 18 minus 12 over 9. 9 Square this you get 9 Square this you get 2. cancel this 2 and make that a 6. so that'll be 18 minus minus 9 is 54. negative 36 but what we care about is that it returns a negative remember that when the second derivative test the second derivative returns a negative that indicates concave down which tells you that the critical value is a Max so this critical value the only one that met all of the criteria when tested in the second derivative yielded a maximum value so our x value or no that's our y value our y value is 3 over square root of two and now we can use our constraint here to figure out our x value so take the 3 over square root of 2 and put it in right here for y and then calculate the x value so 9 minus squared halves we can change that to four and a half nine minus four and a half is equal to four and a half so we have our x value we have our y value and now we can enter them into our main equation and find out what the maximum value for B is so X is equal to 4.5 and Y squared is equal to 9 over 2. calculate that and that will give you your maximum B value and of course four and a half is nine over two so multiplying this we get 81 over 4. so there's your maximized value lifeguard needs to rope off a rectangular swimming area in front of a beach he or she has 1800 yards of rope and floats what are the dimensions of the rectangle that he or she will rope off to maximize the swimming area all right we can say we have a shoreline and then we want to rope off a rectangular area in the water rectangular swimming area it says note the shoreline is one side of the rectangle all right and then if we let this side be X then this side parallel is also X and then this side can be y the first thing we want to do is come up with an equation for area now we know that area of a rectangle is equal to length times width this is going to be our main objective function but right now we have more than two variables we want one input and one output and right now we have two inputs so we need a secondary equation that can help us out here and we can get that by noticing that 1800 yards of rope are going to be used to go around the outside of the area otherwise known as perimeter now typically perimeter of a rectangle is 2x plus 2y however we only need one of the Y's since the other one is accounted for using the shore that means our off to the side constraint equation is 2x plus y equals perimeter and in this case our perimeter is the amount of rope we have and the amount of rope we have is going to be used to cover the distance around the swimming area back over here remember we were trying to rewrite our main equation using only one variable we are going to replace the Y with what it's equal to so getting y by itself here can be done by subtracting 2x from both sides now this can be used as a substitute for y so area equals length times width x times y and in for y I will put all right now I've succeeded in coming up with a main equation an objective function that only uses one input and one output variable which means I can proceed simplify Distributing the X now since the directive here is to maximize I want to go ahead and use calculus so I'm going to find the derivative of the main equation [Music] 1800 minus 4X is good everywhere there are no places where it does not exist so I'll go ahead and set it equal to zero solve for any critical values adding 4X to both sides and then I want to divide both sides by 4. I get an x value of 450 that's one of the dimensions now I need to get the Y value so I can get the Y value by using this equation over here put 450 in for x and that gives you 900 1800 minus 900 is 900. so we have the dimensions last you want to figure out what is that maximum area so area is x times y so go ahead and multiply 450 by 900 and you'll get the maximum area 405 000. and let's see the units here are yards so that would be square yards find the maximum profit and the number of units that must be produced and sold in order to yield the maximum profit assume that Revenue cost of producing X units are all in dollars all right so we are provided with a revenue equation and a cost equation but we're asked about profit so step one we need to get the profit equation and recall that profit is equal to Revenue Minus cost so that's where we'll start 40x minus 0.1 x squared is our revenue revenue Minus cost and then the cost is 5X Plus 20. very important to have the cost in parentheses because it has a minus in front of it and if I distribute that minus it'll be minus 5x minus 20. therefore simplifying this I have 40x minus 5X 0.1 x squared and then minus 20. now if you like you can rearrange that in order of highest to lowest exponents although that's not absolutely necessary to do there's your main equation for profit and now we are looking to again maximize so we want to find the derivative of our main equation we're already good to go because we just have one input X and one output p so we can proceed with our derivative bring down that 2 and multiply it negative 0.2 X plus 35 will be our derivative it exists everywhere so we move on to setting it equal to zero and solving to add 0.2 X to both sides I divide both sides by 0.2 so 175 looks like the number of units that we need we can confirm that that's actually a maximum by using our second derivative so if we go ahead and find our second derivative that second derivative is negative 0.2 which is negative which means concave down and that confirms that our critical value was indeed a maximum 175 is the number of units now to actually find the maximum profit we want to take 175 and put it back into the original equation so put 175 in there and calculate the maximum profit and I hope you are pausing the video to do these calculations the maximum profit with 175 units is three thousand forty two dollars and fifty cents