Problem

A point on the hypotenuse of a triangle is at dis…

Problem 11 Easy Difficulty

A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is $10 \mathrm{~m}$. Find the dimensions of the window to admit maximum light through the whole opening.

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Calculus 1 / AB

NCERT Class 12 Part 1 - Math

Chapter 6

Application of Derivatives

Section 6

Maxima and Minima

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Video Transcript

in this question, a window which is formed in the shape of a rectangle which is surrounded by a semi circles opening. We are. The total parameter of the window is given as 10 m. We have to find the dimension of the window to admit a maximum light through the whole opening. So in this question by the maximum light it is the effort that the surface area. Sorry, the area of the windows should be maximum. Right? So let X me to be the areas of the semi circular opening. Right? So the dimension of the rectangle part would be to X. Multiplied by by sending parameters like this will be the size of the rectangle. Now the parameters of the window which is given a stand meters. The perimeter of the window would be semi circular rock, right? Which is half of Dubai X. Plus the parameters of the rectangle which is equals two, Y plus Y plus two X. Right? So this is the perimeter. And when we simplify this this is given a stand with us. Right? So we get to be a is equal to 10- by X minus two works. Mhm. So why is equal to 10 five plus two X divided by two. Right. So this is the question # one. It is given that the area should be maximum. So let's find out the area of the window area of the window is let's suppose it is dread function. So half of Biaxin square plus the area of the rectangle which is two X. Y. We can substitute the value of buying this equation. Yeah, late. So then minus five plus two X divided by two. So when we simplify this, we get the value of judas one by two minus by X. Square minus four. Ac cisco last 20 X. Right now, since the area of the window should be maximum, so the derivative of the area function should be judo. Mhm. And some about equation when we differentiate this we get -2 paychecks -8 x plus 20. Mhm. Right. So they should be equal to zero from here. We get the from this situation we get the value of access Then divided by by plus four late From a question # one. We can find the value of the by. So the value of rice will be equal to Yeah 10 minus five plus two. Then they were good. Bye bye plus four divided by two. Right? Which is equals two. Then divided by by Plus four. Therefore the length of the window length of the rectangle which will be equal to two X. That is 20 divided by five plus four. The weight of the rectangle is why? Which is equal to 10 by by Plus four. And the areas of the Yeah semi circle is equal to X, which is 10 by pi plus four. Right? So these are the dimensions of the window, right.

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