 Hello and how are you all today? The question says a window is in the form of a rectangle by a semi-circle. Surrounded by a semi-circle. If the total perimeter of the window is 30 meters, find the dimension of the window so that maximum light is admitted. So let us firstly draw a rough figure. This is a rectangle and it is surrounded by a semi-circle. Like this, right? Now here, since this is the diameter, this will be radius, radius and radius. This will be twice of the radius, right? And let us name this figure as A, B, C and D. Now we need to find out the dimensions of this window so that maximum light is admitted. So first of all, we are given the perimeter of the window. Meter. Now, if you carefully observe this perimeter of this window is perimeter of the rectangle plus the circumference of the semi-circle. That is equal to 30 meter. Now I am solving it. We have 4 length plus 4 breadth plus pi r equal to, that implies, now in place of length, we can write 2 r plus 4 breadth, let breadth be x plus pi r equal to 60. This further implies 8 r plus 4x plus pi r is equal to 60. We have 4x is equal to 60 minus, let's take r in the bracket. We are left with 8 minus pi in it. And on dividing the whole equation by 2, we have 2x equal to 30 minus, sorry, it had 2 with it all the way. So now we are left with the bracket in 4 minus pi. Now further, the area of the window is equal to area of the rectangle, that is length into breadth, length is 2 r and breadth is x plus area of the semi-circle, that is pi r square upon 2. Now here in place of x, we can write down 30 minus r pi plus 4 plus pi r square upon 2. Now on simplifying it further, we have it has 60 r minus 2 pi r square minus 8 r square. This further gives us 60 r minus 8 r square. Now on combining these two, we have minus 3 pi r square upon 2. So this is the required area of the window. Now for admitting the area of the window, it is, now a is maximum when dA by dx is equal to 0 and d square A upon dx square is 0. So let us first find out the derivative of the area, that is d by dA of the area, that is 60 r minus 8 r square minus 3 pi r square upon and we will be differentiating it with respect to r, sorry. This implies dA by dr is equal to 60 minus 16 r. This further implies dA upon dr is equal to 60 minus 16 r minus 3 pi. Now on putting this value equal to 0, let us find out the value of r. We have after taking women from here, we are left with in the bracket 16 plus 3 pi. This implies 60 upon 16 plus 3 pi is equal to r. Now let us find out the second derivative. It will be equal to minus 16 minus 3 pi, which is obviously less than 0. So therefore we can say that area of the window when r is equal to 60 upon 16 plus 3 pi. Let us find out the length also then. The length of the rectangle is equal to 2 into r. This gives us 120 upon 16 plus 3, which we took coming out to be in the bracket 4 plus pi upon 2, which is 15 minus substituting the value of r and dividing it by 2. We have 60 upon 2 in the bracket 16 plus 3 pi into 4 plus 16 plus 3 pi. So we have the radius, length and breadth. I just wrote it well. Have a nice day.