Maximizing an Angle In the figure, find the value of c in the interval [0,4] on the x -axis that

Maximizing an Angle In the figure, find the value of c in...

Key Concepts

Calculus-Based Optimization

This concept involves determining the maximum or minimum value of a function using calculus techniques. In problems like maximizing an angle, one typically expresses the angle as a function of a variable, differentiates that function, sets the derivative equal to zero to find critical points, and then verifies whether these points correspond to a maximum or minimum using additional tests such as the second derivative test.

Coordinate Geometry

Coordinate geometry provides the framework for relating algebraic expressions to geometric figures. In optimization problems on the coordinate plane, key elements such as points, lines, and angles are analyzed using their coordinates, enabling the translation of geometric conditions into functions that can be optimized using calculus.

Trigonometric Functions and Angle Properties

Understanding trigonometric functions and the properties of angles is essential when an angle is expressed as a function of a variable. These functions (like sine, cosine, and tangent) relate to the geometric definition of an angle and play a pivotal role in converting geometric problems into algebraic ones, especially when optimizing an expression involving an angle.

Optimization on a Closed Interval

When a function is defined on a closed interval, like [0, 4], optimization techniques must consider not only the critical points where the derivative is zero but also the endpoints of the interval. This ensures that the absolute maximum or minimum is correctly identified, as the optimal value can occur at either a critical point or an endpoint.

Transcript

00:02 So we have this figure here and we're asked to find c such that data is maximum.

00:10 So try to find c such that theta is maximum.

00:24 So that is what we're trying to find.

00:30 We're going to use a few traits from our elementary geometry.

00:35 You know that angles on a straight line adds up to 360.

00:40 You know, there's an angle here.

00:42 There's an angle here.

00:43 They're all subtended by a c.

00:47 Okay.

00:48 So angles on a straight line adds up to 360 degrees.

01:06 So, uh, now.

01:10 Uh...

01:13 No sorry angles on a street line adds up to one eighty degrees that is sixty those is a slip of tongue okay so adds up to 180 degrees so all these three angles are gonna add up to 180 which is pie and we can actually project this one here to form a right triangle so you can have an angle here say a and b here and this is a right triangle and this is also a right triangle.

01:53 And this is also a right triangle.

01:56 And so what is tangent of a? tangent of a is opposite over adjacent.

02:06 Okay, so this is the adjacent, this is adjacent and this is the opposite.

02:11 So what is the length of the opposite? it's just the value of y, which is 2.

02:17 That is the length.

02:18 That is the length of this here.

02:20 Okay, so the value of y is just 2.

02:23 So the length of this side here is 2, actually.

02:29 That is this 2 here.

02:31 So that is the opposite.

02:32 So opposite is 2 over adjacent.

02:36 What is the length over here? it's just 4 minus c.

02:43 That is the length of this portion here, the adjacent.

02:48 So as 4 minus c.

02:51 So that is a tangent of a.

02:53 So a is just going to be inverse tangent of 2 over 4c.

02:59 We can repeat the trick here for b.

03:03 What is the tangent of b? still opposite over adjacent.

03:07 What is the opposite? the opposite is this side, which is still 2.

03:14 And what is the adjacent? the jacent is this side, and that is just c minus 0, which is just c.

03:21 So b is just going to be inverse tangent of 2 over c.

03:32 Okay.

03:34 So that is the two angles here.

03:38 So this angle plus this angle plus this theta here should give it 180 degrees.

03:45 So then, algebraically, theta is going to be 180.

03:50 Which is pi minus b minus a because theta plus b plus a is square pi so if we try to make theta the subject we're going to have this and what is b and a i mean b is inverse tangent of 2 over c and a is is inverse tangent of 2 over 4c.

04:20 So that is what we have here now.

04:24 And we're trying to find c such that theta is maximum.

04:28 And since we're done with maximum values, we're going to find a derivative and set it equal to 0.

04:35 That is our next step because we're trying to find maximum value.

04:42 Okay, so we're going to find, we're going to find the derivative, which is d theta dc, and set it equal to zero.

05:01 So that is what we're going to do next.

05:05 So that is data.

05:06 We're going to rewrite it here.

05:08 Theta is pi minus inverse tangent of inverse.

05:21 Tangent of 2 over c minus inverse tangent again of 2 over 4 minus c.

05:34 So what is d theta d c? we're trying to differentiate.

05:41 This one here with respect to c.

05:44 So this pie here is just going to be 0 because there's no c attached to it.

05:49 And we're still going to use our formula for inverse tangent.

05:54 Which is u prime over uh...

06:03 One plus u squared so what is u here u is just what is in the bracket here so what is u prime u prime is just gonna be negative 2 over c square and over 1 plus what is u u is 2 over c so 2 over c all square over here it's a bit more complicated but we can still do it u is, u here is 2 over 4 minus c.

06:36 So we're trying to find u prime.

06:38 You can be written as 2, 4 minus c to the power of negative 1.

06:44 Okay, and so what is u prime? u prime is just negative 2, 4 minus c to the power of negative 2.

06:56 This is just a function of a function, so you can use a chain rule or whatever.

07:02 Then multiply it by the derivative of what is inside, which is just negative 1.

07:08 So you're just going to have a positive 2 over 4 minus c all squared.

07:17 Because this negative 1 is going to multiply this negative 2, just going to be positive 2.

07:21 And then this one is just going to bring it down to the denominator, okay, because it's 4 minus c 0 to the power negative 2.

07:28 So it can just come down to be like that.

07:30 So that is the derivative of you.

07:33 Is just going to be 2 over 4 minus c, all squared over 1 plus, what is u? u is this guy here.

07:45 So u squared is just going to be that squared.

07:51 Okay, that is the derivative.

07:53 And we can try to clean it up a bit.

07:57 So we just, this negative here is just going to foil out this negative here.

08:02 So it's just going to be 2 over c squared over 1 plus 2 squared is 4 over c squared and minus 2 over 4 minus c squared over 1 plus 2 squared is so 4.

08:21 And then the denominator is also going to be squared.

08:24 Okay, that is what we have so far.

08:27 And what is the gcd between these two terms? gcd, because this is over 1...

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