00:01
Okay, so to maximize the volume here, we need to maximize the cross -sectional area at the end.
00:07
So firstly, we want to know what that cross -sectional area is.
00:11
So by defining this vertical distance here is x, we know we've got a right -angle triangle here with hypotenuse 4.
00:19
So x must be equal to 4 cos theta because it's the adjacent side.
00:26
And this little length up the top, y must be equal to.
00:30
4 sign theta.
00:34
So the area of this trough here, we could divide it into three parts.
00:40
So the area is going to be equal to the central part, which is 6 times x, so 6 times 4 cos theta, plus two end triangles.
00:52
So 2 lots of half times base times height, so half times y times x.
00:58
So half times 4 sine theta times 4 cos theta.
01:06
And so we're left with our area here being equal to 24 cos theta plus 16 sine theta cos theta.
01:22
Or to simplify it slightly, our area is equal to 8 lots of cos theta, all multiplied by 3 plus 2 lots of sine theta.
01:37
Now, to maximise the volume, what we want to do is find the derivative of the area with respect to theta and set it equal to zero.
01:48
So d a by d theta is going to give us using integration, sorry, derivative using the chain rule, it's going to give us eight lots of minus sine theta multiplied by this three plus two sine theta plus 8 cos theta multiplied by the derivative of this term, which is just simply 2 cos theta.
02:19
So setting that equal to 0 and doing a little bit of tidying up, we get that minus 8 lots of sine theta 3 plus 2 sine theta plus 16 cos square theta is going to be equal to 0...