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Problem 34 Hard Difficulty

A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal to the width of the rectangle. See Exercise 1.1.62.) If the perimeter of the window is $ 30 ft $, find the dimensions of the window so that the greatest possible amount of light is admitted.

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Video Transcript

Yeah, okay, here's a norman window, it's a rectangle with a half circle on the top and we know that the perimeter is 30 ft and we want to find the dimensions of the window that lead in the most light. So that means we want to maximize the area. So we need a formula for area of this. It will be easy with the area of a rectangle plus area about half a half a circle. But we've got to name it cleverly. So let's Let's call the radius here to be x radius of the semicircle. So that means this side is two x. And then let's call these sides. Why? Since the vertical? And then if the radius is X, we need the perimeter of the circum half the circumference of a circle. So uh circumference is two pi R of a circle. So in this case circumference is two pi X but we only have half a circle, so one PX Okay, so then the perimeter is going to be this. Why? Plus pi X. That's why plus pi X plus Y plus two X. Okay, so that's our constraint constraint, Y plus pi X plus Y plus two X. Has to Equal 30. Okay, or two Y plus two plus pi X equals 30. And then we're trying to maximize area. So a equals well it's a rectangle. So it's areas linked times with so two X. Y plus the area of half a circle. So remember the area of the circle is pi r squared. So in our case pi X squared and we need half of that. So half the area is half pi X square. Okay, So to find the maximum of anything, you take the derivative and you set it equal to zero and then you solve and then you answer the question Okay, But we have a problem here in our problem is we have too many variables. We have X, Y and X squared. So we need to get one of them out of there and that's what the constraint is for. Okay, so now you decide do you want to solve it for Y or do you want to solve it for X? And then after you do that, which one do you want to plug in? Well, I'm solving for Y. I get to Y equals 30 minus two plus pi X. and so why is 15 -2 Plus pipe X oops. Two plus pi x oops. Mhm. Two plus pi X over two. Mhm. So now my area formula is two times X Times Y. Which is 15 -2 plus pi X over two plus one half by X square. So if I multiply two extra here, I get 30 X plus two plus pi X squared plus one half pi X squared. So I have 30 X plus two X squared plus pi X squared plus one half pi X squared. So one plus a half. That's three halves. So now I have 30 x Plus two x squared plus three halves, pi X squared. All right, That's a so now I'm gonna take the derivative with respect to X. So it's 30 plus for X. I've made a mistake. Right here, That's 30 x two plus pi X squared. And so that is 30 X minus two X squared -909ared. So that is 30 X minus two X squared -1 X half pi X squared. Excuse me? The way I knew that I made a mistake is because I was going to get all positives here and I wasn't gonna be able to set it equal to zero. Alright. Here we go again. Sorry about that. All right. Going to take the derivative derivative of 30 x. 30 minus for X -1/2 pie times two X. So 30 -4 x -9 X equals zero. So 30 equals 4 -9 times X. 04 plus pi So x needs to be 30 over. Okay, sorry four plus pi And then remember we had why was Why was 15 -2 plus pi X over two. So why is 15 two plus pi Over two times 30/4 plus pi. Okay that two cancels with that and give you 15. So now it's 15 times 1 -2 plus pi Over 4-plus pie. I don't know why I'm going crooked there. Okay, so um that's 15. Going to get a common denominator four plus pi That's four plus pi minus parentheses. T two plus pi Over four Plus pipe. So 15 times to over four plus by or 30/4 plus one. Hi so the dimensions are this Okay. X. is 30/4 plus pi. So make the base 60/4 plus by and then make the verticals. Whatever. Why was I forgot already? Um 30/4 plus. Why four Plus by Army. Okay. If you don't like that, change into decimals. Okay so the trick was first you have to get a good constraint and you had before that you had give everything good names. I could have called the base X and then the radius would have been one half X. And then I would have a bunch more fractions and clearly that would not have been good. Okay, so name everything correctly, write your constraint. Figure out what you're going to maximize takes to take the derivative saturday equal to zero. Super fun time.

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James Stewart

Calculus: Early Transcendentals

Catherine Ross

Kayleah Tsai

Caleb Elmore

Kristen Karbon

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Kayleah Tsai

Caleb Elmore

Kristen Karbon

James StewartCalculus: Early Transcendentals

Catherine Ross

Kayleah Tsai

Caleb Elmore

Kristen Karbon

James Stewart

Calculus: Early Transcendentals