00:02
In this problem we have a semicircle of radius 2 and a rectangle that is inscribed in that semicircle.
00:13
So we have a point p in the first quadrant which is on the circle.
00:19
So the coordinates x, y of this point are positive numbers.
00:26
And taking any point on the first quadrant and on the circle we can draw any described rectangle like this.
00:35
So we have one side of the rectangle is all the x -axis, and two points are on the circle.
00:42
And this is symmetry with respect to the y -axis.
00:47
Par a, we want to suppress the area of this in scribe rectangle as a function of x.
00:51
In part b, we do the same for the perimeter of the rectangle.
00:57
In par c, we grab the area of function and search for what value of x is a large as possible.
01:03
And in d, we do the same for the perimeter.
01:06
So let's start with bar a where we express the area function as a function of only x.
01:16
So as we can see on the graph, we have this distance here is x, which is the same like this here.
01:26
Remember, x is a positive number.
01:28
So we have 2x is the width of the rectangle, and the height is y, which is the second core, the positive second coordinate of p.
01:36
So the area of the inscribed rectangle is, let's call it a, and that is 2x times y.
01:57
Remember, 2x is the width of rectangle, and y is the height.
02:03
Now to put that in terms of only x, we need to express y in terms of x, for example, and for that, we remember the point p the coordinate x x and y must satisfy the equation of the circle so the circle has center center at the origin so and radius two so the formula is x square plus y square goes radius square that is four and take into account that y is positive we get we get y is a positive square root of four minus x square and for that reason the area function as a function of x only is 2x times square root of 4 minus x square so this is answer to par a now we do a similar thing with the perimeter of the inscribed rectangle and looking at the sketch we have up here it's called p and we have then one side down here is 2x but we have the same length and the opposite side the parallel side to that so we get 2x and 2x so we get 4x plus and we have y and y that is why both sides the height of the retangle twice so we get plus to y that's it and now we use the expression of y in terms of x we found before and so the perimeter as a function of only x is 4x plus 2 times the square root of 4 minus x square.
03:53
This is the answer for part b.
03:59
And in both cases, in part a and b, the value of x is between 0 and 2 because x is a positive value.
04:10
It can be 0 if we reduce the retangle to a point or line and less than 2 for the square root of 4 minus x squared to be well defined.
04:19
Now in part c, we are going to show the graph of the area function.
04:24
And we have it right here.
04:28
The graph is defined on the interval 0 2.
04:34
So we have 0 here, 2 here.
04:38
And we can see we have sort of vertical tangent line at 2, but that is a critical point, but we know that the largest value of the area function, which is 4 seems to be equal to 4, is happening around 1 .4.
04:54
And that corresponds to it, value in the domain of the function where the derivative is zero because we see clearly here we have a horizontal tiny line.
05:04
So we get to find the derivative of this function area is we have a two constant times product of two functions x and square root of four minus x square.
05:15
We apply the product rule.
05:18
So we have derivative of x is one times four marines x square plus x times x times derivative of the square root of 4 minus x square is 1 half times 4 minus x square to the negative 1 half times the derivative of 4 minus x square is negative 2x so we get that that's equal to 2 times square root of 4 minus x squared then we have a negative because we have a negative here these two cancel out these two factor and so we have negative x square over the square root of 4 minus x square.
06:09
So the derivative of a is two times.
06:15
Now we do the calculation here.
06:19
The common denominator is square root of 4 minus x square times square square root of 4 minus x square is 4 minus x square.
06:29
That's two times 4 minus 2 x square over square root of 4 minus x square.
06:39
We take out the common factor 2.
06:41
So we get 4 times 2 minus x square over square root of 4 minus x square.
06:51
And so this derivative is 0 if and only if the numerator is 0, 4 is different from 0.
06:57
So that can be 0 is 2 minus x squared.
07:03
That's the same as x square equal 2...