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# Optimization: box volume (Part 2)

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

in the last video we were able to get a pretty good sense about how large of an X we should cut out of each corner in order to maximize our volume and we did this graphically what I want to do in this video is use some of our calculus tools to see if we can come up with the same or maybe even a better result so to do that I'm going to have to figure out the critical points of our volume as a function of X and to do that I need to take the derivative of the volume so let me do that and before I even do that it'll simplify things I don't have to use some product rule in some way and then have to simplify that let me just multiply this expression out so let's rewrite volume as a function of X is equal to and I'll write it all in yellow so it's going to be x times I'll multiply these two binomials first so 20 times 30 is 600 then I have 20 times negative 2x which is negative 40 X that I have negative 2x times 30 which is negative 60 X and then I have negative 2x times negative 2x which is positive 4x squared so this part over here simplifies and I can change the order to 4x squared minus 100x minus 100x plus 600 I just switch the order in which I'm writing them so that's that and so I can rewrite the volume of X as being equal to x times all of this business which is let me make sure I have enough space let me into a little higher which is equal to 4x to the third power minus 100 x squared plus 600 X now it'll be pretty straightforward to take the derivative so let's say that V prime of X the prime of X is going to be equal to I just have to use the power rule multiple times so 4 times 3 is 12 X to the 3 minus 1 power 12x squared minus 200 times X to the first power which is just X plus plus 600 and so now we just have to figure out when this is equal to 0 so we have to figure out what 12x weird - 200 X plus 600 is equal to zero what X values get to my derivative to be equal to zero when is my slope equal to zero I could also look for critical points where the derivative is undefined but this turdus derivative is defined especially throughout by the my domain of X that I care about between X is between 0 and 10 so I could try to factor this or try to simplify this a little bit but I'm just going to cut to the chase and try to use the quadratic formula here so this is the X's that satisfy this it's going to be X is going to be equal to so negative B so it's 200 negative negative 200 is positive 200 plus or minus the square root of B squared which is negative 200 squared so I could just write that as 200 squared it doesn't matter if it's negative 200 square root of 200 squared I'm going to get the same value so let me give myself some more space so negative 200 squared well that's going to be 4 with 4 zeros 1 2 3 4 so that's going to be 40,000 minus 4ac so minus 4 times 12 12 times 600 I still didn't give myself enough space times 600 all of that over 2 times a so all of that over 24 and I'll take it take out the calculator again to try to calculate this so let me get out of graphing mode alright so first I'll try when I add when I add the radical so I'm going to get 200 200 plus the square root of 40,000 I could have just written that I could have just written that as 200 squared but that's fine 40000 minus 4 times 12 times 600 times 600 and I get 305 which I then need to divide by 24 which I'll divide by 24 and I get 12 point seven four so one of my possible X's so it equals twelve point seven four and now let me do the situation where I subtract what I had in the radical sign so let me get my calculator back and so now let me do two hundred I probably could have done this slightly more efficiently but this is fine minus the square root of 40,000 one two three minus four times twelve times 600 and I get that's just the numerator and then I'm going to divide that by 24 divide that by 24 and I get three point nine two did I do that right 200 minus 40 thousand minus four times 12 times 600 although all of that divided by 24 my previous answer divided by 24 it gives me three point nine two so it's twelve point seven four or three point nine two now which of these can I use well twelve point x equals twelve point seven four is outside of our outside of our valid values for X if X was equal to twelve point seven four we would cut past we would completely cut Pat the X's would start to overlap with each other so X cannot be twelve point seven four so we get a critical point at X is equal to three point nine two and you could look at the graph that you could say oh look that looks like a maximum value but if you didn't have the graph at your disposal you can then do the second derivative test and say hey our weak concave upwards or concave downwards when X is equal to three point nine two well in order to do the second derivative test you have to figure out what the second derivative is so let's do that the prime prime of X is going to be equal to is going to be equal to 24 X 24 times X to the first minus 200 minus minus 200 and you can just look at inspection that this number right over here is less than four so this thing right over here is going to be less than 100 you subtract you subtract 200 so we can write the second derivative at three point nine two is going to be less zero you can figure out what the net the exact value is if you like so because this is less than 0 we are concave downwards concave downwards another way of saying it is the slope is deke the slope is decreasing the entire time concave downwards when the slope is decreasing the entire time our shape looks like that the slope could start off high lower lower gets to zero even lower lower lower and we even saw that on the graph right over here and since it's concave downwards that implies that our critical point that sits we're doing where the interval is concave downward that critical point is a local maximum is is a maximum so this is the x value at which our function attains our maximum now what what is that maximum value well we could type that back in into our original expression for volume to figure what that is so let's figure out what the volume when we get 23.92 is equal to what is our maximum volumes we get the calculator back out and I could use week it's obviously roughly three point nine two I could use this exact value actually let's well I'll just do three point nine two to get a rough sense of what our maximum value is our maximum volume so it'll be three point nine two three point nine two I'll just use this this expression for the volume as a function of X three point nine two x times twenty minus two times three point nine two times thirty thirty minus two times three point nine two gives us and we need to serve a drumroll now gives us one thousand fifty six point three so one thousand fifty six point three which is a higher volume than we got when we just inspected it graphically we probably could have gotten a little bit more precise if we zoomed in some and then we would a little bit better of an answer but there you have it analytically we were able to actually get an even better answer than we were able to do at least on that first pass graphically