in this video we're going to talk about optimization problems the goal with these types of problems is that you're trying to optimize something that is you're trying to find the dimensions that will maximize the area of a plot of land or that will minimize the amount of fencing required or if you're dealing with business type problems you're trying to find a situation where you can achieve maximum profit or minimize costs and so that's the goal of these type of problems we're trying to find the optimal conditions to get the best results so therefore in order to solve these type of problems you need to be able to identify the maximum and the minimum values of a function so here we have a local maximum and here we have a minimum now there's two there's one important characteristic of these points in the graph at the maximum value and at the minimum value the slope of the horizontal tangent line is zero therefore to find the location of a maximum or a minimum what you need to do is you need to take the first derivative of the function set it equal to zero and solve for x and you're going to see this in action as we work on problems so let's go ahead and do that let's start with this one find two numbers whose sum is 60 and whose product is a maximum so let's put some variables to these two numbers let's say that these two numbers are x and y now we have two variables in order to solve for two variables we need two equations the two equations are going to be the sum and the product so what's the sum of x and y the sum of x and y is simply s is equal to x plus y the product is just what it is it's the multiplication of those two numbers x times y so those are the two equations that we have now here's a question for you which of these two equations would you say is the constraint equation and which one is the objective function the constraint equation is equal to a number a fixed value the objective function it can vary that's what we're trying to maximize or minimize the objective function is the function that we're going to find the first derivative of set it equal to zero and solve for either x or y now we don't know the value of the product but we do know the value of the sum so the sum is a fixed value therefore this is the constraint equation and this is the objective function what we need to do is we need to differentiate p we need to find the first derivative of p set it equal to zero and solve for x or y but right now we can't do that because p is in terms of two variables we need to get p in terms of one variable before we could find the first derivative and that's where the constraint equation comes so what we're going to do is we're going to isolate y in the constraint equation so let's subtract both sides by x 60 minus x is equal to y and then what we're going to do is in the objective function we're going to replace y with 60 minus x so we have p is equal to x times 60 minus x now let's go ahead and distribute x so we have 60x and then minus x squared so now we have p in terms of one variable at this point we could find the first derivative of p and set it equal to zero so p prime is going to be 60 minus 2x and let's set that equal to zero so adding 2x to both sides we get 60 is equal to 2x and then dividing both sides by 2. we can see that x is 60 over 2 which is 30. now at this point we can go ahead and calculate y so let's take 30 and let's plug it into this equation to get y 60 minus 30 is equal to y so y 2 is 30 as well so now we have both numbers whose sum is 60. if we wish we can calculate the product by plugging it into this equation so thirty times thirty we know that three times three is nine and then if we add the two zeros we get a product of nine hundred so that is the maximum product so that's it for this problem the two numbers are 30 and 30 and the maximum product is 900. now let's make sense of these numbers let's see let's check to see if we have the right answer so what we're going to do is we're going to make a table we're going to have x y the sum s and the product p the maximum product occurs when both x and y are 30. we get a maximum product of 900. what if x was 10 and y was 50. in this case the sum is still 60 which that always has to be the case 10 times 50 is 500 so this is not more than 900. if x was 20 y would have to be 40. the sum would still be 60 but 20 times 40 is 800. now let's say if y was 23 x would have to be 37 37 times 23 is 851. it's still less than 900. let's say if x was 32 and y was 28 the sum is still 60. 32 times 28 is 896. or if x is 29 and y is 31 29 times 31 is 8.99 so regardless of what combination of x and y that we choose as long as the sum of x and y is 60. the maximum product that we can achieve is 900. any other combination of x and y with a sum of 60 will be less than 900. so we indeed have the right values of x and y that lead to the maximum product and that's the goal of optimization you want to find the numbers or the conditions that will maximize or minimize something number two find two numbers whose difference is 40 and whose product is a minimum and also find the value of the minimum product so let's write the equations the difference of two numbers let's say the two numbers are x and y so let's say y is larger in value so the difference is going to be y minus x the product just like before is going to be x times y now which will be the constraint equation and which one will be the objective function we know the value of the difference we don't know the value of the product so because we know the value of the difference this here 40 equals y minus x that's going to be the constraint equation this is going to be the objective function the objective is to minimize the product so let's get this equation in terms of one variable let's add x to both sides so we get 40 plus x is equal to y now let's replace y with 40 plus x in the objective function so the product is going to be x times 40 plus x and now let's distribute so here we have x times 40 which is 40x and then x times x which is x squared now let's find the first derivative of the objective function that's going to be 40 plus 2x and then let's set it equal to 0. now i'm going to move the 2x to the other side where it's going to be negative 2x and then i'm going to divide both sides by negative 2. and so x is negative 20. so now that we know the value of x we could plug it into the constraint equation to get y so 40 plus negative 20 will give us y and so that's equal to positive 20. so at this point we have two numbers whose difference is 40 20 minus negative 20 is positive 40. and now we need to calculate the product so it's going to be negative 20 times 20 and so the product is going to be negative 400. so this is the minimum product that we can get from two numbers whose difference is 40. and let's confirm this with a table so we're going to have x y the difference d and then the product p so we had 20 for y and negative 20 for x that gave us a difference of 40 but a product of negative 400. let's see if we can find two of the values for x and y with a difference of 40 but a product that's lower than negative 400. if we were to try 0 and 40 the difference will be 40 but the product 0 times 40 is 0. 0 is higher in value than negative 400 at least on a number line so 0 is not the minimum product if we try a value of 20 for x y it's going to be 40 units higher than x so it's going to be 60 so that the difference is still 40. 20 times 60 is 1200 this is not less than negative 400 is higher we want to find something that's lower than negative 400. if we try 10 and 50 and that's not gonna work this is going to be 500. we were to try let's say negative 10 and 30 30 minus negative 10 that's still positive 40 but this would be negative 300 which is still higher than negative 400 if we were to try negative 25 with 15 the difference is 40 but negative 25 times 15 that's negative three seventy five if we were to try negative twenty one and nineteen we would get negative three ninety nine so the lowest product is negative 400 that is the minimum product that we can get with two numbers that have a difference of 40. and so we can see that the process works so the answer is negative 20 and 20 with a minimum product of negative 400. number three find two positive numbers whose product is 400 and whose sum is the minimum feel free to pause the video if you want to work on this so the product is going to be x times y and the sum is going to be x plus y now we're trying to find we're trying to minimize the sum so the sum is the objective function we have the value of the product so the constraint equation will be 400 is equal to x times y once you replace p with 400 this becomes an equation and so that right there is the constraint equation now let's solve for y so let's divide both sides by x so we get that y is 400 over x now let's replace y with that so the sum is going to be x plus 400 over x so now that we have our objective function in terms of one variable we can go ahead and find the first derivative but before we do that let's rewrite this equation let's move the x variable to the top so this is x plus 400 times x to the minus one now let's find the s prime the first derivative this is going to be 1 minus 400 and then x we'll subtract this by negative 1 so this becomes negative 2. and we'll set this equal to 0. and i'm going to move this value to the other side so it becomes positive so we have 1 is equal to 400 times x and minus 2. now we can put this over one and let's bring the x back to the bottom so that the negative exponent will become positive so this becomes 400 over x squared now let's cross multiply one times 400 is 400 and 1 times x squared is x squared if we take the square root of both sides we get that x is going to be plus or minus 20. now we want two positive numbers so we're going to go with positive 20 instead of negative 20. so that's the value of x that we have so far now let's calculate the value of y so let's plug it into this equation 400 divided by 20 is going to be 20. you can cancel 0 it becomes 40 over 2 which is 20. so we have the two numbers at this point x is 20 and y is 20. now the last thing we could do which is optional we don't have to but we can calculate the sum the minimum sum is going to be 20 plus 20 which is 40. and you can confirm this with a table kind of like what we did before but you'll find that the minimum sum the lowest sum that we can get with two numbers that multiply to 400 is 40. number four find a positive number where the sum of that number and its reciprocal is a minimum so let's use s for sum let's say the positive number is x the reciprocal is one over x now this problem is a little bit different than the other problems we've considered so far because we don't have a constraint equation all we have is the objective function and that's okay because the objective function is not in terms of two variables it's only in terms of one variable so we can go ahead and find the first derivative but before we do that let's rewrite it as x plus x to the negative one let's move this to the top so now let's find the first derivative of s the derivative of x is one the derivative of x to the minus one is going to be we need to move the negative 1 to the front and then we got to subtract the exponent by 1 which makes this negative 2 and we'll set that equal to 0. now let's move this to the other side so we have 1 is equal to one x to the negative two now let's bring this to the bottom so we have one is equal to one over x squared and then we can cross multiply so this gives us 1 is equal to x squared taking the square root of both sides x is equal to plus or minus 1 but we only want the positive number so we're going to go with positive 1. so that's the answer for this problem now let's check our answer so let's make a table where we're going to have x one over x and a sum so when x is one one over x will be one the sum of x and the reciprocal one plus one will be two now can we find some other positive number x where the sum and its reciprocal will be less than 2. remember we want to minimize the sum if we were to make x equal to two this would be one over two two plus a half is two point five that's greater than two not less than it so that doesn't work if x is three this will be one over three this will be about three point 3.33 so let's pick a number lower than 2 but greater than 1. let's say if we chose 1.5 and 1 over x or 1 over 1.5 that's 0.6 repeating and if you add that to 1.5 you're going to get 2.16 repeating which is more than 2. if we're to try 1.1 1 over 1.1 is 0.90 repeating so that's like .9090 and if you add that to 1.1 that will be 2.009 which is more than 2. if we try a number like 0.8 1 over 0.8 is 1.25 and if you add that to 0.8 you get 2.05 so as we can see the minimum sum is going to be two we can't get anything lower than that and that occurs when x and its reciprocal they're both equal to one find the dimensions of a rectangle with a perimeter of 200 feet with an area as large as possible so let's draw a rectangle now let's say this is the length and this is the width the perimeter of a rectangle is 2l plus 2w and we have the perimeter so the perimeter is 200 now to simplify this equation i'm going to divide everything by 2. so this is going to be my constraint 100 is equal to l plus a w the objective function is to maximize the area we want the area to be as large as possible and the area of a rectangle is left times width now let's solve for w in this equation so we need to subtract both sides by l 100 minus l is w and now let's replace w in the objective function with a hundred minus l so we have the area is l times w where w is a hundred minus l now let's distribute l so it's a hundred l minus l squared so now since we have the area of the objective function in terms of one variable only we can now find the first derivative the derivative of a hundred l is a hundred and the derivative of l squared is two l so let's set it equal to zero let's move negative two l to the other side so it can become positive two out and then we could divide by two so half of a hundred is fifty so l is fifteen so now we could use this equation to calculate w so w is 100 minus l so that's 100 minus 50 which is 50. so both the length of the rectangle and the width are equal to 50. now if we want to calculate the area we could simply do length times width and so the maximum area is going to be 50 times 50 and the units is going to be in feet so 50 feet times 50 feet that's 2500 square feet so that's the maximum area for this particular problem and these are the dimensions it's 50 by 50. a farmer has 600 feet of fencing and wants to create a rectangular field along a river he needs no fence along the river itself what are the dimensions of the field that has the largest area so let's begin by drawing the river so here we have the river and he's going to create a fence around this river so there's no need to put a fence here now let's label the dimensions so let's use x and y so let's call this y and let's call this x so the perimeter is going to be x plus 2y instead of 2x plus 2i because there's no x here so the perimeter is just a sum of all three sides so it's x and then y plus y gives us 2y now we have the perimeter the farmer has 600 feet of fencing remember square feet is for area and just feet is perimeter so p is 600 and so this is the constraint now he wants to maximize the area he wants the largest area and the area of a rectangle is left times width or in this case x times y so this is the objective function and this is the constraint for the constraint equation should we solve for x or y i think it's easier if we solve for x because if we solve for y we have to divide by two and you're going to have x over 2 and you have to deal with fractions so let's subtract both sides by 2y or just move it to this side so we have 600 minus 2y is equal to x no fractions so now let's replace x in the objective function with 600 minus 2y so the area is going to be 600 minus 2y times y so let's distribute y so it's going to be 600y minus 2y squared so now that we have the area function or the objective function in terms of one variable we should now find the first derivative so it's going to be 600 minus 2 times 2y which is going to be 4y and then set it equal to 0. now let's move the negative 4y to the other side so it can become positive 4y and then let's divide both sides by 4. so 600 divided by 4 is 150 so we now have the value for y now keep in mind x we said it was 600 minus 2y so that's going to be 600 minus 2 times one fifty two times one fifty is three hundred and six hundred minus three hundred is three hundred so we now have the dimensions y is one hundred fifty feet x is 300 feet now to calculate the maximum area using this equation it's going to be x times y 300 feet times 150 feet and let's see what that's going to be so that's going to be 45 000 square feet so that is the maximum area the dimensions are 300 feet by 150 feet with a maximum area of 45 000 square feet so that's it for this problem now let's move on to the next problem it's similar to the first but a little different so we have this farmer he wishes to create a rectangular field along the river but this time he has a fixed area instead of a fixed perimeter and the area has to be ten thousand square feet what are the dimensions of the field that will require the least amount of fencing and let's also determine the amount of fencing that he needs as well so let's begin by drawing a river and we have the same type of picture so we're going to call this x and y and you can try this problem if you want to as well so this time the area is the constraint so we know the area is x times y and the area is 10 000 square feet now the perimeter is the stuff that we're trying to minimize because we want the least amount of fencing and so that's going to be x plus 2y now i'm going to solve for y in this equation so i need to divide by x so y is going to be 10 000 divided by x so now i'm going to replace y with that so 10 000 times 2 that's 20 000. so the perimeter is going to be x plus twenty thousand and then x to the negative one so now let's find the first derivative that's gonna be one plus twenty thousand times negative one x to the minus two and then let's rewrite the expression so we have one minus twenty thousand over x squared and let's set that equal to zero so i don't need this picture anymore but let's keep in mind that p is x plus 2y so i'm going to take this term move it to this side so i have one is equal to 20 thousand over x squared and i'm going to write one as one over one and then cross multiply so twenty thousand is equal to x squared now let's take the square root of both sides so this will give us the value for x so x if you round it it's going to be about 141.42 feet so now that we have x let's go ahead and calculate y using this equation so y is going to be 10 000 divided by x or divided by 141.42 and so that's going to be about 70.75 so that's the value for y so now that we have x and y which is what we needed to find so that represents the dimensions of the rectangle we can now calculate the minimum perimeter or the minimum amount of fencing that we need so it's going to be x plus 2 times y so the perimeter is going to be approximately 282.8 feet and that's the answer number eight a farmer uses sixteen hundred feet of fencing to enclose a rectangular area which will be divided into three pens what is the maximum total area of the three pens that he can enclose with the limited amount of fencing that he has available well let's draw a picture so we're dealing with a rectangle and this rectangle is going to be divided into three pens now we could call the vertical lengths y and the horizontal length with the horizontal fence in x so that's the situation that we have so far our goal is to maximize the area the area of all three pens is simply x times y its length times width now the total amount of fencing is going to be the length of all of the sides so we have x plus x that's 2x and then we have four y's that we need to add together so that's gonna be plus four y now we can replace f with sixteen hundred so this is our constraint equation now all of the coefficients are even so to simplify this equation i recommend dividing everything by 2. so we get 800 is equal to x plus 2y now it's going to be a lot easier if we isolate x rather than y so let's subtract both sides by 2y so we get that 800 minus 2y is equal to x now let's plug this into our objective equation and so we're going to replace x with what we have here 800 minus 2y so the area is going to be 800 minus 2y times y so we got the area function in terms of one variable now let's distribute y so this is going to be 800 y and then minus 2y squared so now let's find the first derivative the derivative of 800y is just 800 and for negative 2y squared it's negative 4y and we'll set that equal to 0. add in 4y to both sides we get 800 is equal to 4y and dividing both sides by 4 800 divided by 4 is 200 so y is 200 feet now once we have the value for y we can plug it into this equation to get x so 800 minus 2 times 200 that's going to be equal to x so that's 800 minus 400 which is 400 feet so now we have the dimensions that's going to give us the maximum area but the goal in this problem is to simply calculate that maximum area so all we need to do at this point is multiply x by y so x this should be x not y x is 400 feet and y is 200 feet now we know that two times four is eight and then we'll add the four zeros and so the total maximum area is going to be eighty thousand square feet find the point on the line y equals three x plus five that is closest to the origin so let's draw a graph now this linear equation is in slope-intercept form so we could see that m the number in front of x is 3 and b is 5. so the intercept is five and the slope is three so i'm just going to draw a rough sketch and let's say it looks something like that we really don't need an exact picture but somewhere along this line we're going to have a point that is closest to the origin so let's assume it's that point let's call it point p if we draw a right triangle we could say d is the distance between the origin which is represented by o and point p and this is going to be x which is going to be a negative value and y is going to be a positive value so notice that x squared plus y squared is equal to d squared so if we solve for d d is going to be the square root of x squared plus y squared so this is the objective function because we want to minimize the distance and this equation here is the constraint so let's replace y with three x plus five so d is going to be the square root of x squared plus 3x plus 5 squared now we need to find the first derivative of that function so i'm going to rewrite it first so let's rewrite it as this expression and let's raise it to the one half now to differentiate that function we need to use the chain rule so let's start with the power rule let's move the one half to the front keep the inside function the same and then we'll need to subtract the exponent by one one-half minus one is negative one-half and then we need to find the derivative of the inside function the derivative of x squared is 2x and to differentiate this we need to use the chain rule again so it's going to be plus 2 keep the inside part the same 2 minus 1 is 1 times the derivative of the inside function which is 3. so that's the first derivative now let's distribute the one half to everything inside here you shouldn't distribute it to both it doesn't work that way just one of the brackets so one-half times 2x is just x and we can also cancel that too as well because we have to distribute it to everything inside those brackets so on top we're going to have this is d prime by the way x plus i'm going to distribute three to three x plus five so three x times three is nine and then five times three is fifteen and on the bottom if i move this to the bottom the negative exponent will become positive one half and then i can convert this back into a radical so it's gonna be the square root of x squared plus three x plus five squared now we need to set this equal to zero the good thing is if the numerator is equal to zero the entire fraction is equal to zero so let's set the numerator equal to zero x plus nine x is ten x so ten x plus fifteen is equal to zero moving the 15 to the other side we have 10x is equal to negative 15 divided by 10 that becomes 15 is 5 times 3 10 is 5 times 2 so we could cancel 5 so x is negative 3 divided by 2 or negative 1.5 so this is the first answer now we need to find the y coordinate of the point which for the last step is not going to be too difficult to do and so we have y y is 3x plus 5. so that's going to be three times negative three over two plus five so that's negative nine over two and five we can multiply by two over two so it becomes ten over two negative nine plus ten is one so we have the point negative three over two comma one half and so that is the point that is closest to the origin and that's how you can find it number ten find the point on the line y equals four minus x that is closest to the point seven comma six so let's begin by drawing a picture so let's plot y equals 4 minus x so the x intercept if we plug in 0 for x y is going to be 4 so that's actually the y intercept to get the x intercept we need to plug in 0 for y and solve for x so we get an x-intercept of four so we got the point four comma zero and we also have the point zero four so that's a rough sketch of the line y equals four minus x now at the point seven comma six we need to find the point on the line y equals 4 minus x that is closest to this point so how can we do that let's assume that point is here so that point has an x value and it has a y value now let's draw a line connecting these two points we want to minimize the distance between those two points so once again we could use the distance formula d is equal to the square root of x2 minus x1 squared plus y2 minus y1 squared so we could say that 7 is x2 and 6 is y2 and the point that we're looking for we'll call this x1 and y1 so let's replace x2 with 7 and x1 we're going to replace it with x y2 is equal to 6 and y1 we're just going to replace it with y so we have this equation now just so you can understand what's happening here this distance is x this distance is seven if we turn this into a right triangle the base of that right triangle is the difference between 7 and x so this base here is 7 minus x now this part is y and this part here is six the difference between those two parts give us the height of the triangle which is six minus y and so the distance is basically the hypotenuse of that right triangle which is basically the square root of a squared plus b squared so now you see how this equation relates to the pythagorean theorem of a right triangle now there's one more thing we need to do here before we could find the first derivative our goal is to minimize d and right now d is in terms of two variables x and y so what we need to do is we need to replace y with four minus x so d is going to equal now instead of writing the square root symbol we can raise it to the one half so this is going to be seven minus x squared plus six minus and then let's replace y with four minus x squared and all of this is raised to the one-half now six minus four is two negative times negative x that's going to be positive x and so now at this point we have d in terms of one variable we have d as the function of x so now we could find the first derivative and set it equal to zero to solve for x so let's go ahead and do that let's use the power rule let's move the one half to the front we'll keep the inside the same and then we're going to subtract the exponent by one so one half minus one is negative half next based on the chain rule we're going to take the derivative of the inside function so the derivative of 7 minus x squared is going to be 2 times 7 minus x raised to the first power times the derivative of the inside which is negative one for this part it's going to be two times two plus x raised to the first power we got to subtract this by one and then times the derivative of the inside the derivative of two plus x is just one so this part here is going to be the numerator of the fraction that we're about to get but let's distribute one half to these two terms this one half will cancel with the twos that we have here so in the numerator we have 2 plus x and then we have a negative 7 and we have a positive x now this we could bring it to the bottom and it's going to be inside of a square root so on the bottom we have the square root of 7 minus x squared plus 2 plus x squared and we're going to set this equal to 0. now because we have a fraction if the numerator of the fraction is zero the value of the entire fraction is zero so technically all we need to do is set the numerator to zero so here we have x plus x which is two x and then two minus seven that's negative five that's going to be equal to zero adding five to both sides we have two x equals five dividing both sides by two we have our x value which is five over two or two point five so now that we have the value of x we can calculate y y is 4 minus x 4 minus 2.5 is 1.5 1.5 is the same as 3 over 2. so thus we have the point p with the coordinates five over two and three over two so this is the point that is on the line y equals four minus x but it's closest to the point seven comma six so that's our answer number eleven find the point on the curve y equals six x squared minus x cubed plus 10 that has the highest slope and calculate the maximum value of the slope if we want to calculate the slope of the function we need to find the derivative of the function the derivative is a function that gives us the slope at some value of x our goal is to maximize the slope so we need to maximize the derivative function to maximize the derivative function the second derivative must be set equal to zero when the second derivative is equal to zero the first derivative may be a maximum value or it could be a minimum value so that's the goal for this problem we need to find the second derivative of the function and set it equal to zero and then solve for x so let's begin with the first derivative the derivative of x squared is 2x the derivative of x cubed is 3x squared and for the constant 10 the derivative for that is zero so this gives us 12x minus 3x squared now let's find the second derivative the derivative of 12x is simply 12. the derivative of 3x squared is negative 6x if we factor out a 6 we'll get 2 minus x now let's set that equal to 0. we can clearly see that x is equal to 2. now let's make a number line so 2 is the critical point for the first derivative if we were to plug in 3 into this function 2 minus 3 will give us a negative value if we were to plug in 1 2 minus 1 will give us a positive value so to the left of 2 the function is rising to the right of 2 the first derivative not not the original function but the first derivative is decreasing so that's the first derivative has a maximum at two if these signs were reversed that means that it would have a minimum value at two thus the point on the curve with the highest slope occurs at an x value of 2. so now we need to find the y value of that point so let's plug in x into the original equation so we have y is equal to six times two squared minus two raised to the third power plus ten two squared is four times six that gives us 24 2 to the third is eight and 24 minus eight is sixteen sixteen plus ten is twenty six so the point for the first part of the problem occurs at an x value of 2 and a y value of 26 so that's the answer for part a for this problem now let's move on to part b calculate the maximum value of the slope the value of the slope can be found from the derivative function which is right here so we're going to plug in an x value of 2 into this function so it's going to be 12 times 2 minus 3 times 2 squared 12 times 2 is 24 2 squared is 4 times 3 that's 12. so we get a slope of 12. so that is the value of the slope when x is equal to 2. and we could check to see if that is indeed the maximum slope so let's make a table and let's rewrite our first derivative function which was 12x minus 3x squared if we were to plug in 0 the slope would be 0. if we were to plug in 4 this would be 12 times 4 minus 3 times 4 squared that would be 0. if we were to plug in 3 into this equation it's 12 times 3 minus 3 times 3 squared that would be 9. if we're to plug in 1 it will be 12 minus 3 which is also 9. so we could see that the maximum slope occurs at an x value of 2. so that's it for part b so part b is equal to 12. part a the answer is 2 comma 26. number 12 a rectangle is inscribed in a semicircle with a radius of 10 centimeters what are the dimensions of the rectangle that will maximize its area and also calculate the maximum area well let's draw a picture so let's say this is the semi-circle let's see if i can do a better drawing and we have a rectangle inscribed inside of the semi-circle so here is the y-axis and this is the x-axis now the radius of the semi-circle is 10. so that means this point here along the x-axis is 10 and this point along the y-axis is 10. now let's call this part of the rectangle x and let's call this part y so drawing a rectangle like this we can see that the length of the rectangle is x and x so it's 2x and the width of the rectangle is this portion which is y so the area of the rectangle is length times width the length is 2x the width is y so the area is just 2x times y now what is y in terms of x to answer that question we need to know the equation of a circle the equation of a circle is x squared plus y squared is equal to r squared solving for y squared if we subtract both sides by x squared we get that y squared is r squared minus x squared and then if we were to take the square root of both sides y is going to be plus or minus the square root of r squared minus x squared so that's the equation for a full circle now we have two signs positive and negative for the upper half of the circle it's positive root r squared minus x squared for the bottom half of the circle it's y equals negative r squared minus x squared so since we're dealing with the upper half of the semi-circle or rather the upper half of the circle which is a semi-circle we could say that y is just the positive square root of r squared minus x squared and r is 10. so r squared is going to be a hundred so we have y is equal to the square root of 100 minus x squared so we can say that the area is going to be 2x times the square root of 100 minus x squared the area function is the objective function that's the function that we're trying to maximize and right now we have it in terms of one variable so we can go ahead and find the first derivative and set it equal to zero now we need to use the product rule since we basically have two things multiplied to each other and if you recall the derivative of f times g is f prime g plus f of g prime so think of the 2x as f and the square root of 100 minus x squared sg so let's go ahead and find the first derivative of the area function so the derivative of 2x is simply 2 that's f prime we're going to keep g the same so that's the square root of 100 minus x squared next we'll keep the first part of the function the same 2x and then we need to differentiate this so keep in mind this is 100 minus x squared raised to the one half so this is going to be times one half we'll keep the inside part the same we'll subtract the exponent by one one half minus one is negative a half and then we need to differentiate the inside part of the function the derivative of negative x squared will be negative 2x now let's take a minute to make some space so i'm going to rewrite the area function here it's l times w which is 2x times y which is 2x times the square root of 100 minus x squared so now let's delete this and so we have a prime is equal to 2 square root 100 minus x squared now we can cancel the two and the one half and so we're left with x times negative 2x which is going to be negative 2x squared and this term we can move it to the bottom make this exponent positive and convert it back into its radical form so on the bottom we're just going to have the square root of a hundred minus x squared now this we can put it over one what we need to do is we need to get common denominators so we can combine it into a single fraction so i'm going to multiply this fraction by 100 minus x squared top and bottom so now we have a prime is two when we multiply these two the square root will disappear and we're just going to get a hundred minus x squared and then we have the other term so at this point we can write it as a single fraction so we can say that a prime this is 2 times 100 which is 200 here we have negative 2x squared plus another negative 2x squared so that's going to be negative 4x squared all over the square root of 100 minus x squared so we're going to set this equal to zero if we multiply both sides of this equation by a hundred minus x squared these will cancel on the right we'll get zero times the square root of 100 minus x squared which is still zero so the end result is that we'll have the numerator equal to zero if we add four x squared to both sides we get two hundred is equal to four x squared dividing both sides by four we get 50 is equal to x squared and then we could take the square root of both sides the square root of 50 we can write that as the square root of 25 times the square root of 2 and the square root of 25 is 5. so now we know what the value of x is so x is 5 square root 2. so now we can calculate y we know that y is the square root of 100 minus x squared which is what we have here and we know that x squared is 50. if you square this you'll get 50. so this is the square root of 100 minus 50 which is the square root of 50 and we know that's going to simplify to 5 square root 2. so in this problem both x and y have the same value now our goal in part a is to determine the dimensions of the rectangle that will maximize this area which means we want to calculate the length and the width of the rectangle so the left of the rectangle is 2x and x is 5 root 2. so the length of the rectangle in this problem is going to be 10 square root 2. so that's the first part of the answer the width of the rectangle is simply equal to y and y is 5 square root 2 and the units are centimeters so we can write the dimensions as 10 square root 2 by 5 square root 2 left times width now part b calculate the maximum area now that we know the dimensions of the rectangle we can calculate the maximum area that we can form when the rectangle is inscribed in a semicircle with a radius of 10 centimeters so the area is simply going to be the length of the rectangle times its width the length is 10 square root 2 the width is 5 square root 2. now 10 times 5 is 50. and the square root of two times the square root of two is the square root of four which is two so it's fifty times two which is a hundred for these types of problems whenever you have a rectangle inscribed in a semicircle and you're given the radius a quick way to find the maximum area of the rectangle that's inscribed in such a semi-circle is you could use this formula the maximum area is simply going to be r squared so in this problem r is ten ten squared is a hundred so you get a hundred square centimeters so don't forget to pay attention to the units so that's a quick and simple way to calculate the maximum area it's always going to be r squared for this type of problem so the units for the dimension will be the same as the unit for the radius but the unit for area will be square units so this is inches the area will be square inches if this was feet the area will be square feet you