I have a question. Why can’t the area just be xy. Like would it still work. I tried it with radius 4 and I am getting 8 as my area. I don’t get why it has to be (2x) (2y) is there something else to do instead of splitting it into fourths
Hi! i have a question, why r^2 is a constant for this example? i saw time rates with the radius getting derived at, isn't it the same with optimization?
I also struggled with understanding why r is a constant. I think its helpful to consider that r is NOT a function of x and y. Because the rectangle is inscribed, by definition, we know r will not change even though there are infinite possibilities for x and y. Because of this, we treat it like a constant when deriving dA/dx
Why would you need to do all that? A square is a rectangle. You just have to draw a line (radius) from the center, let's call this A. Then, another line (radius) at 90° of the first line, Let's call this B. Now, you already have 2 sides of a triangle, just calculate the hypothenus using Pythagorean Theorem using the 2 sides above, sqrt(A^2+B^2). Because the 2 sides (A and B) are equal, same as the length of the radius (r), you can simplify this by doing sqrt(2r^2). Or you can alternatively, you can use the sine function, r/sin 45°, since the resulting triangle would be an isosceles triangle and you have a given 90° on 1 angle. The result would be 1 side of the rectangle. Since it is a square, then you just multiply it by itself. In simplest form, it would look like this, Area=sqrt(2r^2)^2 Alternatively, Area=(r/sin 45°)^2
your accent helps me pay attention i have no idea why
This video is going to save my final grade!! This solution makes sense now!!
Glad it helped. Good luck in your class.
I have a homework question with the exact same same question lol, this was really helpful
calculus early transcendentals section 8th edition 4.7 question #25
@@gartyqam It's a very common Calculus book for college.
Midterm exam tomorrow, I really appreciate this video
Good luck on your exam.
Love your voice and this explanation is really easy to follow
big thumbs up
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Sir I am new to your channel and your explanation amazing it is clear as a crystal of water
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How did the r disappear at the end? Shouldn’t it be r-r/2^1/2?
it's like 1-1/2 = 1/2
I have a question. Why can’t the area just be xy. Like would it still work. I tried it with radius 4 and I am getting 8 as my area.
I don’t get why it has to be (2x) (2y) is there something else to do instead of splitting it into fourths
Because center of circle is at the origin.
@@crowsmathclass so is it necessary to split it up. My teacher taught us to have the area as xy
Yes. Because we center at origin.
@@crowsmathclass is there a situation where we can just use A= xy without 2x 2y
Like how would we know if it is centered and not centered
so are we solving for 2x at the end rather than singular x or y?? i was bit confused.
x
Thankyou sir!. This was pretty helpful.
You’re welcome. Glad it helped.
How would you find the radius of the entire circle?
The radius is a variable. Hypothetically, you can plug any number to it.
Can you please help me
So i was thinking what if the tye rectangle has a length of 11 cm and width of 6 cm.
How do we solve that ??
just input the numbers into x and y I think
Thank you, it really help me understand the problem.
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Very helpful, thank you!
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I LOVE YOUR ACCENT
Thank you.
How can you say that area is maximum
As proof, use the 1st or 2nd derivative test. I think it will show you that that is indeed the maximum area
Hi! i have a question, why r^2 is a constant for this example? i saw time rates with the radius getting derived at, isn't it the same with optimization?
We are looking for the dimensions of the rectangle. r is a constant in this problem. So we are looking for x and y.
I also struggled with understanding why r is a constant. I think its helpful to consider that r is NOT a function of x and y. Because the rectangle is inscribed, by definition, we know r will not change even though there are infinite possibilities for x and y. Because of this, we treat it like a constant when deriving dA/dx
Shouldn't the second term for the derivative of the area be -4x^2/sqrt(r^2-x^2) instead of -4x/sqrt(r^2-x^2)? -x*4x= -4x^2.
He did input it there unless I'm missing something? His answer is right.
Super super helpful
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i love your accent
Thank you.
Thanks man. nice voice
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Why would you need to do all that?
A square is a rectangle.
You just have to draw a line (radius) from the center, let's call this A. Then, another line (radius) at 90° of the first line, Let's call this B. Now, you already have 2 sides of a triangle, just calculate the hypothenus using Pythagorean Theorem using the 2 sides above, sqrt(A^2+B^2). Because the 2 sides (A and B) are equal, same as the length of the radius (r), you can simplify this by doing sqrt(2r^2).
Or you can alternatively, you can use the sine function, r/sin 45°, since the resulting triangle would be an isosceles triangle and you have a given 90° on 1 angle.
The result would be 1 side of the rectangle. Since it is a square, then you just multiply it by itself.
In simplest form, it would look like this,
Area=sqrt(2r^2)^2
Alternatively,
Area=(r/sin 45°)^2
You should find second derivative and find Local maximum and minimum for area,which can give surety for maximum area
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Thank u so much
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My god thank u so much
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u should go faster
Sir I am new to your channel and your explanation amazing it is clear as a crystal of water
Thank you sir
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