hi everyone today we're going to talk about how to find the dimensions of a rectangle
with a given perimeter 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 dimensions let's take a look in this particular optimization
problem we've been asked to find the dimensions of a rectangle that have a perimeter of one
hundred meters whose area is as large as possible so as with any optimization problem the first
thing we want to do is draw a picture of what we've been asked to find so that we can visual
what we're doing so we have a picture of a rectangle we know since we're dealing with
a rectangle that two of the side will be equal we can call then x the other two sides will
also be equal so we'll go ahead and call them y obviously we know immediately then that
the area of this rectangle will length times width so area will x y and we know that the
perimeter is one hundred meters so that means that the sum of all four sides will have to
be equal to one hundred meters so we can say that the perimeter of one hundred meters will
be equal to two x plus two y so that's pretty much everything we know from the problem we've
been given now that we have a picture and an idea of where we're headed we need to identify
a constraint equation and an optimization equation which you'll always need to do when
you're dealing with an optimization problem so your constraint equation is the limitations
within which you've been asked to work so we've been asked here to find the dimensions
of a rectangle and we know we want to maximize the area but we've been told that we have
to work within the constraint that the perimeter is one hundred meters so we can do anything
we want but the perimeter has to be one hundred meters since that dimension constrains us
this is our constraint equation the optimization equation will be the equation that we want
to either maximize or minimize in this case we're looking for the largest possible area
that means we're trying to maximize the area or the area equation so a equals x y will
be our optimization equation whichever equation it is you're trying to maximize or minimize
will be your optimization equation so once we've identify the constraint equation and
the optimization equation what you need to do is determine whether or not you have your
optimization equation in one variable in our case we still have x and y in our optimization
equation and we really need the optimization equation in one variable only before we can
move forward so what we'll do is we'll solve the constraint equation for one of our variables
plug that into our optimization equation so we get it in terms of one variable and then
move forward let's go ahead and solve the constraint equation we can solve it for either
x or y doesn't matter it will work either way but let's go ahead and solve it for y
so we can get our optimization equation in terms of x we'll subtract two x from both
sides to get one hundred minus two x equals two y and then we'll divide both sides by
two to get y equals one hundred minus two x all over two when we divide through by two
because we can simplify the two in the denominator will cancel and we'll just get fifty minus
x is equal to y now we can plug this value in for y into our area equation to get area
equals x times fifty minus x now that we have our area equation in terms of one variable
only we want to simplify as much as possible take the derivative and set it equal to zero
in order to take the derivative it will help us if we distribute the x so let's go ahead
and say the area is equal to fifty x minus x squared that's pretty much as simple as
we can get it now we want to take the derivative we'll call the derivative a prime and when
we take the derivative we'll get fifty minus two x now that we have the derivative function
and it's as simple as it can be we can go ahead and set this equal to zero and solve
for x so zero equals fifty minus two x if we add two x to both sides we'll get two x
equals fifty and if we divide both sides by two we'll get x equals twenty five now a lot
of people since they've solved for x are attempted to stop here but this is the point in which
we already need to return to our original problem to remember what we're actually being
asked to find we've been asked to find the dimensions of a rectangle with an area as
large as possible find the dimensions so we need to find the dimensions x and y we already
have the x dimension we need to go and find the y dimension so here we have an equation
for y in terms of x we can go ahead and plug x in so we'll get y equals fifty minus twenty
five which is what we just found and we'll get y equals twenty five so in this case both
our x and y dimensions turned out to be the same and we can see that the rectangle with
an area as large as possible is actually a square so what we have is a square a really
bad square with dimensions twenty five by twenty five and we can check ourselves because
we can see that the area will be six hundred and twenty five which seems big enough and
the perimeter we know if we have twenty five all the way around twenty five times four
will give us a perimeter of twenty five times four which we know is one hundred which is
the constraint we were given in the original problem so that's it those are the dimensions
twenty five by twenty five twenty five by twenty five that maximize the area of the
rectangle so I hope you found that video helpful if you did like this video down below and
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