If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Optimization: sum of squares

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.B (LO)
,
FUN‑4.B.1 (EK)
,
FUN‑4.C (LO)
,
FUN‑4.C.1 (EK)

## Video transcript

we are being asked what is the smallest this is a little typo here what is the smallest possible sum of squares of two numbers if their product is negative 16 so let's say that these two numbers are x and y x and y so how could we define the sum of the squares of the two numbers so I'll just call that the sum of the squares s4 sum of the squares and it would just be equal to x squared plus y squared and this is what we want to minimize we want to minimize minimize s now right now s is expressed as a function of x and y we don't know how to minimize with respect to two variables so we have to get this in terms of only one variable and lucky for us they give us another piece of information their product is negative 16 so x x times y is equal to negative 16 so let's say we wanted this this expression right over here only in terms of X well then we could solve we can figure out what Y is in terms of X and then substitute so let's do that right over here if we divide both sides by X we get Y is equal to negative 16 over X and so let's replace our Y in this expression with negative 16 over X so then we would get our our sum of squares as a function of X is going to be equal to x squared plus y squared Y is negative 16 over X negative 16 over X and then that's what we will now square so this is equal to x squared plus what is this 256 256 over x squared or we could write that as 256 256 X to the negative 2 power that is the sum of our squares that we now want to minimize well to minimize this we would want to look at the critical points of this which is where the derivative is either zero or undefined and see whether those critical points are possibly a minimum or a maximum point they don't have to be but those are the ones if we have a minimum or maximum point they're going to be one of the critical points so let's take the derivative so the derivative s Prime we do this in a different color as prime of X I'll do it right over here actually the derivative s prime of X with respect to X is going to be equal to 2x times negative 2 times 5 6 so 2 2 X plus 256 times negative 2 so that's minus 512 X to the negative 3 power x to the negative 3 power now this is going to be undefined this is going to be undefined when X is equal to 0 but if X is equal to 0 then Y is undefined so this whole thing breaks down so that isn't a useful critical point x equals 0 so let's think about any other ones well the its defined everywhere else so let's think about where the derivative is equal to 0 so when does this thing equal 0 so when does 2x minus 5 12 X to the negative 3 equals 0 well we can add 5 12 X to the negative 3 to both sides so you get 2x is equal to 512 X to the negative third power we can multiply both sides times X to the 3rd power multiply both sides times X to the 3rd so all the X's go away on the right hand side so you get 2 X to the fourth is equal to 512 we can divide both sides by 2 and you get X to the fourth power is equal to 256 and so what is the 4th root of 256 well we could take the square root of both sides just to help us here so let's see if we take so it's going to be x squared is going to be equal to 256 is 16 squared so this is 16 this is going to be x squared is equal to 16 or X is equal X is equal to 4 now that's our only critical point we have so that's probably the x value that minimizes our sum of squares right over here but let's make sure it's a minimum value and to do that we can just do our second derivative test so let's figure out let's take the second derivative as prime prime of X and figure out if we are concave upwards or downwards when X is equal to four so s prime prime of X is going to be equal to two and then we have negative three times negative five twelve so I'll just write that as plus 3 times 512 that's going to be 1500 and 36 is that right yeah 3 times 500 is 1500 3 times 12 is 36 X to the negative 4 power so this thing is going to be and this thing right over here is actually going to be positive for any X this thing is this thing is going to be positive for any X X to the negative 4 if even if it's a negative x value that's going to be positive everything else is positive this thing is always positive so we are always in a concave concave upwards situation concave upwards means that our graph might look something like that which I don't want to draw a little squiggle it might look something like that and you see the reason why the second derivative implies concave upwards the second derivative positive means that our derivative is is constantly increasing so the derivative is constantly increasing it's negative less negative even less negative let me do it in a different color you see it's negative less negative even less negative 0 positive more positive so it's increasing over the entire place so if you have a critical point where the derivative is equal is equal to 0 so the slope is equal to 0 and it's concave upwards you see pretty clearly that we have minimized we have minimized the function so what is y going to be equal to we actually don't even have to figure out what Y has to be equal to in order to minimize the sum of squares we could just put it back into this but just for fun we see that Y would be negative 16 over X so Y would be equal to negative 4 and we could just figure out now what our sum of squares is ours minimum sum of squares is going to be equal to it's going to be equal to 4 squared which is 16 plus negative 4 squared plus another 16 which is equal to 32 now I know some of you might be thinking hey I could have done this without calculus I could have just try out numbers whose product is negative 16 and I probably would have tried out 4 and negative 4 in not too much time and then I would have been able to maybe figure out that it's lower than if I did 2 and negative 8 or negative 2 and 8 or 1 and 16 and that's true you probably would have been able to do that but you still wouldn't have been able to feel good that that was a minimum value because you wouldn't have tried out for point 0 1 or 4 point 0 0 1 1 in fact you couldn't have tried out all of the possible values remember we didn't say that this is only integers it just happened to be that our values worked out to be just worked out to be integers in this situation you can imagine what would happen if the problem wasn't if their product is negative 16 but what if their product is negative 17 or what if their product is negative 16 point 5 or what if their product was PI squared then you wouldn't be able to try everything else out and you would have to resort to doing what we did in this video