Honeycomb The surface area of a cell in a honeycomb is S=6 h s+(3 s^2)/(2)((√(3)-cosθ)/(sinθ)) w

Honeycomb The surface area of a cell in a honeycomb is S=6 h...

Key Concepts

Differentiation

Differentiation is the process of computing the derivative of a function, which measures its rate of change. In optimization problems, differentiation is used to find where the function's slope is zero, indicating potential maximum or minimum points.

Optimization

Optimization involves finding the maximum or minimum values of a function under certain conditions. In calculus, this typically requires setting the derivative equal to zero to identify critical points and then testing these points along with the boundaries of the domain to determine the optimal value.

Trigonometric Functions

Trigonometric functions such as sine and cosine play a key role in problems involving angles. Their well-defined properties and derivatives allow for the analysis of functions that incorporate these terms, making it possible to study the behavior of functions involving angles.

Critical Points

Critical points are the values at which a function's derivative equals zero or is undefined. These points are crucial in determining local minima or maxima in optimization problems, and assessing their nature often involves additional tests or consideration of the function’s behavior at the endpoints of its domain.

Domain Restrictions

Domain restrictions specify the set of valid input values for a function. When optimizing a function, it is important to consider these restrictions, as the optimal solution must lie within the given interval. This is particularly important for physically meaningful problems where parameters, such as angles, are bounded.

Transcript

00:01 So they tell us that the surface area of a honeycomb cell is given by this equation here.

00:06 And they say hs are constants and theta is changing.

00:11 And what they want us to do is figure out what angle theta minimizes our surface area.

00:16 So maybe off on the side i should probably write h .s.

00:21 Are equal to constants.

00:27 So now, remember any time we want to minimize or maximize something, what we want to do is take the derivative of it, because taking the derivative of it will help us find the critical values or where our horizontal lines are going to, or horizontal tangents are going to exist, which, where the horizontal tangents exist are normally where our maximums and our minimums are.

00:51 And then, since we're on this interval, the extreme value theorem is going to tell us that we also need to check those as well.

00:59 So what we're going to do is first take the derivative of this, set equal to zero, find all the values in between pi 6, pi half, that satisfy that.

01:08 And then we're going to go ahead and also check the endpoints to just see which one minimizes this overall.

01:18 So let's go ahead and do that.

01:21 So if i take the derivative of this with respect to theta.

01:27 So the 6hs, that's just a constant, so that goes away.

01:31 So we have d -theta is equal to.

01:33 And that 3s squared over 2 constant, so that just gets pulled out when we do the derivative.

01:39 And that we have d by the theta of root 3 minus cosine theta over sine theta.

01:47 So to take the derivative of this, we can use quotient rule.

01:51 So let's do that.

01:54 So the quotient rule, the way i always have to write this out so i don't.

01:58 Don't do it incorrectly.

02:00 So d by d x of high over low.

02:06 So this is low d high d high minus and let me scoot this over a little bit.

02:16 So low d high minus high d low all over the square of what is below.

02:23 So that's just a little jingle i have memorized for that.

02:26 So we're going to do just bring sign up so it'll be sine of theta.

02:33 And now we're going to take the derivative of what's in the numerator.

02:36 So the derivative of root 3 is going to be 0.

02:39 The derivative of cosine is negative sine, but that negative there cancels out.

02:44 So that's just going to be sine theta.

02:47 And then we do minus.

02:49 And actually, i'll need to probably scoot this over.

02:54 All right, i'll just scoot it down.

02:57 So, yeah, now high.

02:59 So it would be root 3 minus cosine theta.

03:03 And then we take the derivative of sine, so that's going to just be cosine.

03:08 And then this is going to be all over sine squared theta.

03:14 Now, let's go ahead and do a little bit of algebra to simplify this down.

03:18 Or better yet, i don't think we need to actually find what this derivative is, if i'm mistaken.

03:26 Yeah, so since we don't need to find the derivative, let's just set this equal to zero right now.

03:31 And then we can start solving for this.

03:33 Alright, so we can multiply over that 3 sine squared over 2.

03:38 Just get that away.

03:41 And doing that, we will end up with, so sine squared theta.

03:49 And then if we distribute the cosine and the negative, that would give us negative root 3, cosine theta, and then plus cosine squared theta, all over sine squared.

04:05 Theta and so this is still equal to zero now you might recall that sine square theta plus cosine squared theta that is just going to simplify down to one so we can rewrite this as one minus root 3 cosine theta over sine squared theta is equal to zero now to find our possible maxes and mince remember we would just want to solve this here but we also want to check for our derivative is going to be undefined.

04:39 So other possible is just going to be where our denominator is equal to zero.

04:44 So sine square of theta is equal to zero or sine of theta is equal to zero.

04:49 But notice we don't actually need to worry about this because our original function is not going to be defined where sine of theta is equal to zero.

04:57 So we can kind of just toss these out.

04:59 But just kind of keep in mind those are important things to kind of remember.

05:02 All right.

05:03 So we have that.

05:04 Now what we want to do is go ahead and, and just multiply over by sine squared.

05:11 So that gives us 1 minus root 3.

05:13 Cosine theta is equal to 0.

05:16 And then we can add that over...