hi everyone today we're going to talk about how to find the area of the largest rectangle
that will fit inside a semicircle to complete this problem we'll draw a picture of the problem
and write what we know identify optimization and constraint equations and then use the
derivative of the optimization equation to find the area let's take a look in this particular
problem we've been asked to find the area of the largest rectangle that can be inscribed
in a semicircle that has radius r so as with any optimization problem you guys know the
first thing that we need to do is draw a picture of what we've been asked to find and write
down all of the information that we can with some types of optimization problems like this
one remember that it can helpful to draw the picture on an x y coordinate plane so I've
drawn the x and y axis here and then drawn my semicircle and made it symmetrical about
the y axis so I have my semicircle and we know that the semicircle has a radius of r
so that means that at any point wherever it meets the x axis here we have a radius of
r over here on the left we know that we're out a distance negative r from the origin
so I can go ahead and write negative r we'd also be a distance of r up here on the y axis
because obviously the radius is the same everywhere so we've drawn that information in and then
we've shown a rectangle that's inscribed inside of the semicircle now here's where we start
assigning arbitrary values so let's pretend that from the origin we go out a distance
of x to get the edge of the rectangle here so we say we're out a distance of x and let's
say that the height of the rectangle here is y if that's the case then that means that
this point here right where the rectangle meets the semicircle is at the coordinate
x y because we're out a distance of x and we're up a distance of y that means also that
where the rectangle meets the semicircle over here on the left our coordinate will be negative
x y so again we're just getting a visual of what this rectangle looks like so now we can
start getting an idea of the dimensions of the rectangle because we know the height of
the rectangle is y right and again we don't know what y is yet but we know the height
is y and we know that the length of the rectangle is two x right because this length here is
x and this length here is x so if we want the full length of the rectangle we know that
that's two x so that means that we can write the area equation for our rectangle of two
x the length times y the height so area is length times height or two x y we also know
that the equation of a circle is x squared plus y squared equals r or the radius squared
this should look familiar remember that when we're talking about the unit circle the radius
is one we have a radius of one from the origin so one squared stills gives us one and you
should be familiar with the equation x squared plus y squared equals one that's the equation
of the unit circle in this case we don't know what the radius is so we just have to call
it r which means the equation of the circle is x squared plus y squared equals r squared
so now that we have everything written down we need to identify a constraint equation
and an optimization equation whenever you have optimization problems like this you'll
always need a constraint equation and you'll always need an optimization equation the optimization
equation should be the obvious one because it's what you're trying to optimize so in
our in our problem in our question we've been asked to find the area of the largest rectangle
what they're asking us for is to maximize the area of the rectangle since we're maximizing
area area the equation for area will be our optimization equation so to give you another
example if you're asked to you know minimize surface area of something surface area the
equation for surface area will be your optimization equation so in this case a equals two x y
will be our optimization equation our constraint equation is to defined by whatever we're constrained
by or whatever we're given in the problem whatever limits us we know the only thing
we've been told is that the radius of the semicircle must be r so whichever equation
involves r is going to be our constraint equation we are constrained within the limits of this
equation here x squared plus y squared equals r squared so this will be our constraint equation
and what we need to do after we've identified the optimization and constraint equations
is to make sure that we have our optimization equation in one variable in this case unfortunately
we have it in two variables x and y we need to solve our constraint equation for one of
those variables so that we can plug in to our optimization equation and get our optimization
equation in terms of only one variable so let's go ahead and solve our constraint equation
either for x or y in this case we'll solve it for y doesn't matter it will work either
way but we'll solve it for y keep in mind that in this problem r is not a variable it's
a constant because the radius is constant everywhere right and this problem could've
have just as easily asked us to find the area of the largest rectangle that could be inscribed
in a semicircle of radius four or eighteen right whatever the value of r is it's always
going to stay the same so we can treat it as a constant x and y are our only variables
so let's go ahead and solve this constraint equation for y we'll subtract x squared from
both sides and we'll get y squared equals r squared minus x squared if we take the square
root of both sides to get y on its own we'll get y equals the square root of r squared
minus x squared now that we've solved for y we can go ahead and plug this value the
right hand side into our area equation for y so we'll get area equals two x times y which
is the square root of r squared minus x squared now that we've plugged we want to make sure
our area equation or our optimization equation is as simple as possible and once its as simplified
as it can be we want to take the derivative of our optimization equation we can't really
simplify this any further so we want and go ahead and take the derivative which we'll
call a prime once we take the derivative we're going to set it equal to zero and solve for
x so a prime notice that we'll need to use the product rule we have two functions here
two x being one of them the other one being the square root of r squared minus x squared
so we're going to treat those as two separate function and use the product rule to take
the derivative remember that the product rule tells use to take the derivative of the first
function so the derivative of two x we know is two and then multiply by the second function
without changing anything so r squared minus x squared square root of r squared minus x
squared then we're going to add to that the opposite case so instead before we took the
derivative two x now we're going and leave that alone and this we're going to take the
derivative of the square root of r squared minus x squared so we can treat this r squared
minus x squared instead of the square root this is the same thing as r squared minus
x squared to the one half power as is any square root so when we take the derivative
of this we'll use chain rule remember that chain rule tells us to take the derivative
of the outside function first leaving the inside function completely untouched and then
multiply by the derivative of the inside function so this here r square minus x square is our
inside function the outside function is the whole thing including the one half if we ignore
our inside function and pretend its something simple and we take the derivative of the outside
function we're going to bring this one half out in front here multiply by one half so
we'll one half times now we want to leave the inside function completely alone so r
squared minus x squared and then subtract one from the exponent as we do with any derivative
so we'll get negative one half and then we want to multiply according to chain rule now
we need to multiply by the derivative of the inside function so we're multiplying by the
derivative of r square minus x squared we'll remember that r squared is a constant the
derivative of any constant is zero so the derivative of that will be zero and the derivative
of negative x squared will just be negative two x so we multiply by negative two x now
we need to simplify this and then set it equal to zero and solve for x so here we'll get
two times the square root of r squared minus x squared and then over here on the right
we'll get this two and this two to cancel we'll just have this negative two x which
we'll move out in front and combine with this x here to get negative two x squared and then
all we're left with is the quantity r squared minus x squared to the negative one half if
we move that whole term to the denominator this exponent here will turn from negative
one half into positive one half which is the same thing as the square root so we'll get
r squared minus x squared or the square root of r squared minus x squared here in the denominator
to continue simplifying we'll go ahead and find a common denominator in order to find
a common denominator we'll multiply this first term times the square root of r squared minus
x squared divided by the square root of r squared minus x squared so we'll have two
times the square root of r squared minus x squared this first term we'll multiply it
by the square root of r squared minus x squared divided by the square root of r squared minus
x squared this is the same as multiplying by one because this fraction simplifies to
one minus two x squared over the square root of r squared minus x squared now we see that
we have a common denominator the denominator of the first term is same as the denominator
of the second term so we can combine the fractions what we'll get is a prime equals we'll have
two times these two terms here will combine and the square roots will go away so we'll
be left with r squared minus x squared minus the two x squared here so minus two x squared
all divided by our common denominator the square root of r squared minus x squared when
I simplify the numerator I'll get two r squared minus two x squared minus two x squared all
divided by the denominator which we'll save some and denote as d and now we want to combine
the two x squares here so we'll two r squared minus four x squared all divided by the denominator
and then let's go ahead and factor out a two and bring our denominator back so two times
r squared minus two x squared all over the denominator of the square root of r squared
minus x squared now what we want to do is since we've simplified this as much as possible
is set the derivative equal to zero so we've set the derivative equal to zero and we want
to solve for x the only time when our when our fraction over here on the right hand side
is going to equal zero is when the numerator equals zero when the denominator equals zero
of course the function is undefined but we want the function to equal zero which means
we have to get the numerator to equal zero so if we just look at the numerator for now
we can divide both sides by two and we'll get zero equals just r squared minus two x
squared we need to solve this for x in terms of r so we'll go ahead and add two x squared
to both sides we'll divide by two to get r squared over two and then we'll take the square
root of both sides to solve for x so when we take the square root of both sides we can
deal with the numerator and denominator separately the square root of r squared is just r and
then we have the square root of two here in the denominator we can ignore the negative
obviously we would have positive or negative r over the square root of two in this case
because we took a square root but if had the negative term we'd have a negative radius
and we can't have a negative radius because we're dealing with an actual figure in three
d space and a negative radius would mean the semicircle and rectangle didn't exist so we
can just deal with the positive term so we know now that x is equal to r divided by the
square root of two remember though that what we really care about is what we've been asked
to find in the problem find the area of the largest rectangle so in order to find the
area of the rectangle not only do we need the x dimension we also need the y dimension
and then we need to plug those into our area equation to find the area so in order to find
the y dimension we'll plug our x coordinate back into our equation for y and we'll get
the square root of r squared minus r over the square root of two squared and we'll simplify
this to get the square root of r squared minus r squared over two to find a common denominator
we'll multiply the first term by two over two and we'll get two r squared over two when
we combine fractions we'll get two r squared minus r squared just gives us r squared divided
by two in the denominator when we take the square root here we'll get r in the numerator
and the square root of two in the denominator notice that it's the same value that we got
for the x coordinate so the y and x coordinates are actually equal to one another and again
we can ignore the negative term we get obviously positive or negative r over the square root
of two but we can ignore the negative term because y had to be a positive value and our
last step of course is just to find the area because that's what we've actually been asked
for so we'll get the area is two times the x coordinate which we know to be or the x
value which r over the square root of two times the y value which is also y over the
square root of two and when we simplify we get two r squared over two our twos will cancel
and we'll see that the area is equal to r squared this tells us that the area of the
largest rectangle that can be inscribed in a semicircle that has radius r is r squared
now since we have this equation we can plug in any value for the radius r and automatically
get the area of the largest rectangle so if we're ever told that the radius is four we
know immediately that the are of the largest rectangle will sixteen if we're told that
the radius is ten we automatically know that the area of the largest rectangle will be
one hundred so we can use this equation with any value of r that we're given so I hep you
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