The trough in the figure is to be made to the dimensions shown. Only the angle θcan be varied. W

The trough in the figure is to be made to the dimensions...

Key Concepts

Geometric Modeling

Geometric modeling is the process of translating a physical scenario or design into a mathematical framework. This involves identifying relevant dimensions and relationships—in this case related to volume and angles—and expressing them as mathematical functions that can be analyzed and optimized using techniques like differentiation.

Trigonometric Function Optimization

Many optimization problems, particularly those involving angles, rely on trigonometric functions such as sine and cosine. Optimizing such functions involves understanding how variations in the angle affect the overall function and applying differentiation techniques to find the angle that yields an optimal (maximum or minimum) value.

Calculus-based Optimization

This concept involves applying calculus techniques, especially differentiation, to find the maximum or minimum values of a function. In optimization problems, you typically set the derivative equal to zero to find critical points, which may correspond to maximum or minimum values of the function under consideration.

Transcript

00:01 So for us to be able to find the maximum volume of this trough, well, let's first just go ahead and write what our trough equations volume is.

00:11 So it would be the volume is, well, it's this base here times the length of it because it's just like we stretched it going all the way across.

00:20 So it would be 20 times whatever the base is area is.

00:26 And now let's go ahead and try to figure this out.

00:28 Well, here we at least know that from here to here, this is all.

00:34 Of length 1 because it's some kind of a rectangle.

00:37 I don't know if we necessarily know this is a square yet.

00:42 So what we can do instead to try to find this area is what we need to know the height of these triangles and then also the base of these triangles, this little b.

00:56 And then once we have that, we can go ahead and get the area of this using the tropsoid formula.

01:03 So this should be equal to, so i'll write this off down here, so capital b is one half, so base one plus base two, all over two, divided by its height.

01:22 Well, base one, we can say is just one, and then base two is this up top here.

01:32 So if we were to just kind of plug all that in, that's one half.

01:38 1 plus, so it would be 2b plus 1, all over 2, and then we don't know what our height is yet.

01:48 Oh, actually, if i have this one half out here, i shouldn't need the 2 dividing there.

01:54 So yeah, i'll just leave it like this here.

01:58 Now we can go ahead and simplify that because that would just be b plus 1 times h.

02:05 All right, so now we need to figure out, well, what is b and what is h in terms of theta? like, that would help us out a lot.

02:11 Well, since here we have a right triangle and we have this angle theta, we know the angle directly across from this is going to be the hypotenuse of this times theta.

02:23 So this is going to be 1 sine theta.

02:27 So that's going to be our little b.

02:29 And then our height here is supposed to be adjacent.

02:33 So this is going to be 1 times cosine theta.

02:39 So we can go ahead and plug those in.

02:41 So our base is sine of theta plus one and our height is cosine of theta so now we have our base so let's go ahead and plug that in up here into our volume equation so our volume is going to be 20 times cosine theta times one plus sine theta so that's the same thing i just rewrote it a little bit so now let's go ahead and just just repeat the cosine so that would be 20 cosine theta plus cosine theta sine theta there should be 20 here also so this is our volume equation now if we want to try to maximize this remember we take the derivative of this so we get dv by d theta is equal to well we take the derivative of cosine which gives us negative signs this would be negative 20 sine theta plus 20 times well, cosine times sine, we need to use product rule.

03:51 So we're going to have cosine theta times the derivative of sine, which is going to be cosine theta.

03:59 And then over here we take the derivative of cosine.

04:03 So then that's going to give us negative sign.

04:05 So it would be negative sine theta times sine theta.

04:09 And then let's go ahead and simplify this down a little bit.

04:13 So it'll be minus 2.

04:16 Sine theta plus 20 cosine squared theta minus 20 sine squared theta and then this is going to be equal to well we want this to be equal to zero because this is how we go about maximizing something.

04:40 Now in order for us to solve this well we would want to rewrite this using some trig identities and the one we're going to want to use is one of the pythagoras ones.

04:50 So what i'm going to use is that cosine squared is equal to 1 minus sine squared.

04:56 So let me just go ahead and write that down here.

04:59 So negative 20.

05:02 And actually, before we do that, let's just divide everything by 20, just to make this look a little bit prettier for us.

05:10 So doing that, we would end up with negative sine theta plus cosine squared theta.

05:19 Minus sine squared theta is equal to zero...

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