How big can y be?

here we've got a nice geometry problem that we're going to solve using first semester differential calculus so our setup is that we've got two squares with side length one so here's one over here and then there's one right here that's been tilted a little bit and our goal is to find the maximum value of this quantity y right here so i want to notice that this y is constructed by finding the line from this vertex of the square to this vertex of the other square and then that intersection up here with this vertical line which intersects this kind of leftmost vertex of the tilted square so first off notice that the minimum value of y is obviously one and we get that by rotating the square until its bottom side is in line with this lower line okay so let's see what we need to do in order to find this maximum value so we're going to put this in the coordinate plane so i'm going to say that this coordinate right here is x and so it's along the x-axis at well so as well so this will be the coordinate x comma zero okay great and then since this has side length one we know this coordinate over here will be x plus one comma zero and then again this is side length one so we know this coordinate right here is x plus one comma one and now we can really keep going so we know this coordinate right here is x comma one now next i'm going to introduce an angle that we will eventually eliminate from the picture but we'll need it for a little bit i'm going to call it theta and it'll be this angle right here so notice if that is theta that makes this distance right here sine theta so that makes this coordinate right here x comma sine theta and i'll put the sine theta in blue because we'll eventually like i said eliminate the dependence on theta so that everything will be in terms of x okay so now let's see what we've got over here so these two angles are going to be complementary and that's because theta plus 90 degrees plus whatever this angle needs to be 180 so i'll write that as pi over 2 minus theta and then likewise this also has measure angle theta kind of for the same reason so that tells us that we know this coordinate down here so this will be sine theta comma zero and again i put the sine theta in blue because we'll eventually like i said replace that with something in terms of x so now similarly we know this coordinate up here will be zero comma cosine theta [Music] furthermore we know that the length of this line segment is cosine theta giving us the identity that x equals sine theta plus cosine theta so that's actually going to be pretty important okay so let's see if we can fill in anything else so we'll add a couple of line segments in so one line segment will come down from here so that it would be perpendicular to the x-axis and connect off with this thing here okay good so now i want to notice that since this is angle theta this is pi over 2 minus theta that's going to make this thing right here also equal to theta for the same reason that we had described before over here okay and that also makes this here angle theta okay nice now next we could set this bit equal to a so that hypotenuse that makes this distance right here a sine theta then we can set this bit equal to b that makes this bit b sine theta and then furthermore we know a plus b equals 1 so that means the distance from here i'll put this in orange to here i'll put that in orange as well is sine theta we know this coordinate is x comma zero that makes this coordinate right here x minus sine theta comma one but by that formula up there we know that x minus sine theta is cosine theta so let's maybe put that here so we've got this coordinate is cosine theta and comma 1. now next we'll also complete a triangle over here by taking a horizontal line in this direction noticing that this angle is complementary to theta so it's pi over 2 minus theta so that tells us that this height right here is equal to sine theta again because we see that this angle here is theta but that makes this coordinate 0 comma cosine theta plus sine theta so we know that cosine theta plus sine theta is x so that makes this coordinate right here zero comma x and then while we're at it we'll fill in this coordinate right here which is zero comma y okay great so now those are all of the points that we need and now what we'd like to do is take this formula right here and somehow solve for sine theta and or cosine theta so that we can eliminate the sines and the cosines from here and everything will be in terms of x so let's maybe see how we could do that we'll take this formula and then we'll square it that'll give us x squared equals sine squared theta plus 2 sine theta cos theta plus cos squared theta we know sine squared plus cosine squared equals 1 we can move that over leaving us with x squared minus 1 equals 2 sine theta cosine theta we'll replace cosine theta with x minus sine theta giving us 2 times sine theta and then x minus sine theta but now looking at this carefully we see that we have a quadratic equation where sine theta is like the variable that we can use to solve for sine theta in terms of x so let's maybe write that down uh carefully so we've got 2 sine squared theta so that comes from 2 sine theta times sine theta and then minus 2 x sine theta so that comes from 2 sine theta times x and then i've moved those to the other side of the equation that's why the signs changed and then we have plus x squared minus 1 equals 0. so now we can use the quadratic formula on this and we'll see that sine theta will be equal to the following expression in terms of x it is one half and then x plus the square root of 2 minus x squared so we've got that as the value of sine theta and then cosine theta we know is x minus sine theta which means we can easily get an expression for cosine theta as well so let's see what it is so that is going to be one half x minus the square root of two minus x squared so now what we'll do is take this value of sine theta in terms of x this value of cosine theta in terms of x and plug it in everywhere here and we're about ready to go on to the next step on the last board we introduced an angle into our picture we called it theta and then we expressed lots of points in the plane in terms of that angle finally we were able to express sine of that angle and cosine of that angle in terms of x where x comma 0 is this coordinate right here so i filled everything in from that analysis okay so next what we'll do is calculate the slope of this line here two different ways so we'll first calculate it using this point and this point and then next we'll calculate it using this point and this point and then that'll give us an equation for y in terms of x that we can maximize like i said before using first semester calculus okay so first taking the slope between this line and this line we'll get one minus y over x plus one so like i said that's going to be this slope right here now next we'll take the slope between this point and this point and let's see what we get so we'll get change of y over change of x so let's see that'll be equal to one minus x over x plus one minus one half times x minus the square root of x well square root of two minus x squared so we've got something like that so now we need to go about simplifying this and solving for y so maybe first off let's multiply by a minus 1 on both sides that'll cancel this to y minus 1 and cancel this to x minus 1. next we can multiply by x plus 1 giving us y minus 1 equals x squared minus 1 over well now we need to simplify that denominator so we have x minus half x so that's going to be half x plus 1 and then plus half times the square root of 2 minus x squared okay nice so now we can add 1 to both sides and we have y equals 1 plus x squared minus 1 over 1 half x plus 1 plus 1 half square root of 2 minus x squared now next maybe what we'd like to do is multiply the numerator and the denominator of this second term by two just to make it look a little nicer and then we're ready to jump into the calculus part so let's see we've got y equals one plus two x squared minus two over x plus 2 plus square root of 2 minus x squared like that okay so now let's bring that to the top and we'll discuss the next steps okay so i brought our formula for y to the top of the board and it's kind of gnarly but if we just go through it step by step it shouldn't be too bad so like i said we want to maximize the value of y that's our whole goal we're going to use differential calculus so that means we need to take the derivative in order to find the critical points those critical points will allow us to find the maximum so here we'll use just the quotient rule so we'll have y prime equals we don't need to worry about this one because it's a constant the derivative of a constant is zero so the derivative of the numerator 4 x times the denominator x plus 2 plus 2 minus x squared to the half and then from that we need to subtract the derivative of the denominator times the numerator so let's see the derivative of the denominator will be one plus one half times two minus x squared to the minus half but then we're going to need to multiply by negative 2x because of this minus x squared inside so that's going to leave me with a minus x here and then like i said we need to multiply by the numerator which is 2x squared minus 2 and then all of this is over the denominator which is x plus 2 plus 2 minus x squared to the half so we want to set this derivative equal to zero to find the critical points but then we're subtracting two big objects in the numerator so those need to be equal in order for us to get zero in the numerator those two big things i'll overline in red so it's this guy and this guy so instead of setting that numerator equal to zero what we really can do is just erase the denominator and change that minus sign to an equal sign and then we've got an equation to solve okay so let's maybe do that okay so upon our discussion from the last board we have this equation to solve so let's maybe expand this a little bit we'll have 4x squared plus 8x plus 4x times 2 minus x squared to the half so that's our left hand side and then our right hand side will be 2x squared minus 2 and then plus 2x minus 2x cubed times 2 minus x squared to the minus half and now we're at the point where this is not super fun to solve by hand so it's most definitely possible to solve by hand but you could also plug it into wolfram alpha or mathematica or something so i'll just jump to the solution what you get here is x equals one half and then you'll have minus one plus the square root of five and then next it'll be minus the square root of 2 times the square root of 5 minus 2. so in the end you can reduce this to a quartic polynomial and then you can solve that quartic polynomial to get this value for x but that's the value of x that we need our goal was to find the maximum value of y so you can take this value of x and plug it into y and again after some reduction which is really just kind of a pain but not really too complicated you can get the following value so you'll get y equals 1 plus the square root of 10 times the square root of 5 minus 22 and so that's our maximum value of y and that's a good place to stop

Channel: Michael Penn

Views: 60,297

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn

Id: yiyL7MeanR4

Length: 14min 55sec (895 seconds)

Published: Sun Feb 07 2021