Metal rain gutters A rain gutter is made from sheets of metal ...

Question

Metal rain gutters A rain gutter is made from sheets of metal 9 in wide. The gutters have a 3-in base and two 3-in sides, folded up at an angle Θ (see figure). What angle Θ maximizes the cross-sectional area of the gutter?

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says an engineer is designing a V-shaped water channel to maximize flow capacity. The channel is made up from a 12-inch wide sheet of metal with the sides bent upwards at an angle theta. Each side of the channel, when bent, measures 6 inches. What angle theta maximizes the cross-sectional area of the channel for water flow, express your answer in degrees. Well, the problem we're given a diagram that contains our V shape for our channel, which each side of the V is labeled with 6 inches. We're also given the angle theta, which is the angle measured between the horizontal and the right side of the channel. We're also given 4 possible choices as our answers. For choice A, we have 15 degrees, for choice B, we have 30 degrees, for choice C, we have 45 degrees, and for choice D, we have 60 degrees. Now, the question wants us to find the angle theta that maximizes the cross-sectional area of the channel. So, before we do that, we're going to actually label a couple quantities in our diagram. The first thing we're going to do is bisect our channel, and this is going to create a right triangle. Where the height of this right triangle is going to be labeled as H. And the horizontal leg of this right triangle, it's going to be labeled as B divided by 2, where B here is going to be the total um base of our channel. So, here we have half of the channel, so, this distance is going to be B divided by 2. Now, the reason why we drew this triangle, right triangle, is because we can draw the same right triangle on the outside of the channel. But this right triangle is now going to contain our angle theta, where the leg of the right triangle that is adjacent to the angle theta has a length of B divided by 2, and the leg of the triangle that is opposite of the angle theta has a length of H. So, we need to find the angle theta that maximizes the cross-sectional area. So we need to determine the cross sectional area of our channel. So, here, our cross-sectional area is going to be that of a triangle. So recall your formula for the area of the triangle that our area A is going to be equal to one half the base of the triangle, which we will label as B, multiplied by the height of the triangle, which is H. So, what we want to do is relate B and H to our angle theta, since that's what we're the parameter that we're trying to maximize the cross sectional area with respect to. So, here we can actually relate B to the angle theta using trigonometry. So, since B divided by 2 is the layer of the triangle that is adjacent to the angle theta, we will have cosine of theta as being equal to. The adjacent lake, which is B divided by 2. Divided by the hypotenuse of this right triangle, which is 6 inches, the length of one of the sides of our channel. So we'll divide this by 6. So we can actually solve this expression for B by multiplying both sides of our equation here by 12. In doing so, we'll get B is equal to 12 multiplied by cosine of theta. We can also relate H to our angle theta, and here we'll use the sine functions, so we'll have the sign of theta, and that's going to be equal to H divided by 6, since the sine of data is equal to the opposite leg divided by the hypotenuse, which in this case is H divided by 6. So we can solve this expression for H by multiplying both sides by 6, and so we'll have H is equal to. 6 multiplied by sign of data. So, we can substitute both of these quantities back into our formula for the area. So this is going to be equal to. One half, multiplied by the quantity of 12 multiplied by cosine of theta in quantity, multiplied by the quantity of 6 multiplied by sin of theta in quantity. So, simplifying this, this is going to be equal to. 36 Multiplied by sine of theta, multiplied by cosine of theta. Now we can combine those two trick functions using our double angle formulas. So recall that the sine of 2 theta is equal to 2 multiplied by sin of theta multiplied by cosine of data. So this is going to be equal to 18, multiplied by the sine of 2 theta. So now what we want to do is find the critical points for our area function. So this means that we're going to need to calculate the derivative of our area with respect to theta. We'll calculate the derivative of A with respect to theta, that's going to be equal to the derivative with respect to theta of the quantity of 18 multiplied by the sine of 2 theta. So taking this derivative, this is going to be equal to 18, multiplied by, and when I take the derivative of a sine function, I get a cosine function, so we'll have 18 multiplied by the cosine. Of 2 theta, but here we have to apply our chain rules, so we need to take the derivative of 2 theta with respect to theta, and so that's gonna be 2, so we need to multiply our cosine function here by 2. So simplifying this expression, this is going to be equal to 36 multiplied by cosine of to theta. So we called to find critical points, we need to set our first derivative here equal to 0 and to solve for theta. So we'll have 36. Multiplied by cosine of 2 theta. is equal to 0. So, dividing both sides by 36, we'll get cosine of your theta is equal to 0. Now, cosine function is zero when its argument is equal to 90 degrees or minus 90 degrees. But here we're only interested in angles that are going to be between 0 and 90 degrees. So, what we're going to do is set our argument here, which is 2 theta equal to 90 degrees. And so dividing both sides by 2, we get theta is equal to 45 degrees. This here is our critical point. So we need to determine whether this is a maximum or not. So, to determine this, we're gonna need to take another derivative. So, we're going to be looking at the second derivative of our area with respect to data. And this is going to be equal to the derivative with respect to theta. Of the quantity of 36 multiplied by the cosine. Of Tohada In quantity. So taking this derivative, this is going to be equal to 36. Multiplied by the quantity, and when I take the derivative of the cosine function, I'll get a minus sign function, so we'll have minus sign. Of 2 theta. In quantity, and again, we'll have to apply our chain rules, so we'll need to multiply this by the derivative of the argument, which in this case is 2. So this is going to be equal to minus 72, multiplied by sin of to theta. So now we need to evaluate the second round of when data is equal to 45. So, we'll be looking at the second round of our area with respect to theta. Evaluated at theta is equal to 45 degrees. So this is going to be equal to minus 72, multiplied by the sign of the quantity of 2 multiplied by 45 degrees. In quantity And so this is going to be equal to. 72, that the sign of 90 degrees is equal to 1. And so, therefore, I'm just left with -72, and this value is negative, so it is less than 0. And so recall that if your second derivative here is negative, then that means that our function here is going to be concave down. And if our function is concave down, that means that our critical point here is actually a maximum. So, The theta equal to 45 degrees is the angle that maximizes our area. So that is the answer that we're looking for. And when we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice C. So I hope that this explanation has been useful, and I'll see you in the next video.

Metal rain gutters A rain gutter is made from sheets of metal 9 in wide. The gutters have a 3-in base and two 3-in sides, folded up at an angle Θ (see figure). What angle Θ maximizes the cross-sectional area of the gutter? <IMAGE>

Key Concepts

Cross-Sectional Area

Optimization

Trigonometric Functions

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