We want to determine the dimensions of a rectangle bounded by the semicircle y equals the square root
of the quantity sixteen minus x squared that would maximize the area. So here's an example of what
the rectangle might look like. Let's see if we can determine the dimensions of this rectangle.
The length from here to here would be equal to y, but y is bounded by the function,
so the length of this side here would be equal to the square root of
the quantity sixteen minus x squared. Now looking at the horizontal distance here,
the distance from the origin to this corner would be x, and therefore the length on the
other side would also be equal to x. So the length of this entire side would be two x.
The dimensions for this rectangle are two x by the square root of the quantity sixteen minus
x squared, which means the area of this rectangle would be the product of these two dimensions. So
we would have two x times the square root of the quantity sixteen minus x squared.
Now we want to maximize this area, which means we'll have to determine
the critical numbers of the function by determining where the derivative of
this function is either equal to zero or does not exist. Looking at this function,
we'll have to apply the product rule in order to find the first derivative.
So we'll let this equal the first function, and we'll let this equal the second function.
The derivative is going to be equal to the first function, which is two x,
times the derivative of the second function, which is going to require the chain rule. So
the inner function will be sixteen minus x squared; we'll call that u.
We can think of this as u to the one half power. The derivative of u to the one half power with
respect to x would be one half times u to the negative one half power, but u is
sixteen minus x squared, times the derivative of u, which is negative two x. Then we have the
second function, which is the square root of the quantity sixteen minus x squared,
times the derivative of the first function, which is two.
Now let's work on simplifying this. The two in the numerator and the two in the denominator would
simplify out, so it looks like we would have negative two x squared, and then this
has a negative exponent, so we'll move it down to the denominator. We'll have
the quantity sixteen minus x squared to the one half power, plus here we're just going
to have two times the quantity sixteen minus x squared to the one half power.
We'll go ahead and put this over one because what we're going to do now is combine these
two fractions. We're going to do that by multiplying by the least common denominator,
which would be the quantity sixteen minus x squared to the one half power. So we multiply
this fraction by the quantity sixteen minus x squared to the one half power over itself.
When we do this, notice how we now have a common denominator;
we have sixteen minus x squared to the one half power in the denominator. Our
new numerator is going to be negative two x squared plus two times the quantity
sixteen minus x squared. Now we can distribute here and combine like terms.
So the first derivative is going to be equal to negative two x squared minus two x squared plus
thirty-two, all over our denominator. Now we want to know where the derivative does not
exist or where it's equal to zero. Looking at the denominator, we can see it's not going to
exist when x is equal to four. If x is equal to four, the height of the rectangle would be zero,
so we can disregard that and only determine where this is equal to zero.
This is only going to equal zero when the numerator is equal to zero. So if we set
the numerator equal to zero and try to solve, let's factor out negative four. We're left
with x squared minus eight equals zero. This will equal zero when x squared equals eight.
So we can take the square root of both sides. The two critical numbers are x equals positive
or negative two times the square root of two. So these are the two critical numbers,
but because one of the sides has a length of two x, we know x can't be negative;
otherwise, that would produce a negative area.
So the only critical number that we have is
x equals two times the square root of two. This means that at this x value,
there may be a maximum or minimum function value, which does represent our area.
There's a couple of ways of doing that; we could either use the first derivative
test or the second derivative test, but looking at our first derivative here,
I really don't want to determine the second derivative of this because it
would take some time. So for this example, I think we'll just look at it graphically.
But what we're hoping to determine is that to the
left of two times the square root of two, the function will be increasing,
and then to the right of two times the square root of two, we hope the function will be decreasing,
so that verifies we'd have a relative maximum at x equals two times the square root of two.
Let's determine the decimal approximation for two times the square root of two. It's
approximately two point eight. I've already entered in the original area function,
so let's go and take a look at the graph, and notice that about two point eight, right in here,
we do appear to have a relative maximum. The function is increasing and then decreasing.
Therefore, we're going to go ahead and assume that at x equals
two times the square root of two, we will have a relative maximum. So let's go ahead
and determine the dimensions of this rectangle as well as the maximum area.
To do that, remember the dimensions were two x by the square root of the quantity sixteen minus x
squared. So let's evaluate these two expressions. If x is two times the square root of two, we'd
have two times two times the square root of two, which would be four times the square root of two.
Then for the square root of the quantity sixteen minus x squared, we'd have sixteen
minus two times the square root of two times two, which equals eight, so sixteen minus eight equals
eight. So we have the height of eight, which is two times the square root of two.
Therefore, the dimensions of the rectangle that would maximize the area would be
four times the square root of two units by two times the square root of two units,
which means the maximum area would be the product of these two,
which would be four times two equals eight times two equals sixteen square units.