Light transmission A window consists of a rectangular pane of ...

Question

Light transmission A window consists of a rectangular pane of clear glass surmounted by a semicircular pane of tinted glass. The clear glass transmits twice as much light per unit of surface area as the tinted glass. Of all such windows with a fixed perimeter P, what are the dimensions of the window that transmits the most light?

Accepted Answer

Welcome back, everyone. A garden plot is to be fenced off and divided into two equal areas by a partition parallel to its width. If the total amount of fencing available is 100 m, including the partition, what length L and width W should the garden plot have to maximize the area. So let's understand what's happening in this problem. While we have a rectangular area. And we are essentially dividing it into two equal parts, right? So let's make sure that we have our partition roughly in the middle. And now we have to understand that we are going to have 3 widths. And now if we look at the top and the bottom, well those are represented by length L. So we have 100 m of fencing available for those 3 widths and 2 lengths. So we can set a constraint such that 2 L plus 3W is equal to 100 m, right? So we cannot exceed that, and that's our constraint. What we're trying to do is basically. Maximize the area. What is the area? Well, essentially it is length multiplied by width right for the whole rectangle. So area is equal to L multiplied by W, length multiplied by width, and what we want to do is simply perform optimization. But since we have two variables, we're going to use our constraint to express one in terms of the other. So we can just go for width. We're going to subtract 2 L from both sides, 100 minus 2 L, and then we have to divide both sides of the equation by 3 to express width in terms of length and therefore our area. As a function of length becomes length multiplied by width which is now expressed by. 100 minus 2 L divided by 3. From here we can essentially distribute the terms, so let's go ahead and do that. First of all, let's factor out 1/3. And then we're going to have 100 L. Minus 2 L squad. So that's our area function, and now we're going to differentiate it. So we're going to find the first derivative of the area function with respect to length. We can factor out the constant and now we want to differentiate 100l, which is simply 100. Based on the power rule and the derivative of L square there's 2 L, so we end up with 100 minus 4L because we have 2 multiplied by 2 L. Well then, we have successfully applied the power rule, and now we want to set this derivative equal to 0 because we are interested in the critical points. Notice that it is in a partially factored form, so we can set 100 minus 4 L equals 0, and if we solve this equation, well L is going to be equal to 25. Now from here, we essentially want to prove that this point is a point of maximum, and we can see that using the 2nd derivative test. So if we find the second derivative of area with respect to land. This gives us. 13, we still have our constant and now in the derivative of 100 to 0 and the derivative of -4 L is simply -4. So 4/3, right? It basically tells us that our function is concave down. For any l, including the critical point, and this implies that we indeed have a local maximum which is represented by the critical point. So we have proved that l equals 25 gives us a maximum. So we have determined the required length and now for width. We have to recall that width is 100 minus 2 L, so 2 multiplied by 25 divided by 3. Let's evaluate the result. We end up with 50. Divided by 3. Well then, so we have our answers. Now we're going to state them with units. So I length. That maximizes the area is L equals 25 m and width equals 50 divided by 3 m. Those would be our final answers for this problem. Thank you for watching.

Key Concepts

Optimization

Perimeter Constraint

Area and Light Transmission

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