Fidn the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius 6cm

- anonymous

Fidn the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius 6cm

- katieb

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- anonymous

- anonymous

you're looking for the area of a square with a diagonal length of 12. |dw:1433820528039:dw|

- anonymous

ok how would i start , use both areas?

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- anonymous

- anonymous

Calculus?

- anonymous

yes

- anonymous

- anonymous

Well, peach was correct about what type of figure you'd need, but I assume you have to show it, lol.
Okay, so optimization. We need a way to relate information about the circle with the rectangle inscribed in it. So we need the formula for what we want to optimize, which is the area of a rectangle.
Area of rectangle = LW
Next we want some known value of the circle, whether it be from a formula or just one of its dimensions, that we can somehow relate to the length and width of the rectangle.
So we know the radius of the circle is 6. Which means the diameter of the circle is 12. Can you think of a way to relate the radius or diameter of the circle to the length or width of the rectangle?

- anonymous

um one side of rectanlge is 12cm long ?

- anonymous

Well, that wouldnt be possible actually. 12 is the length of the diameter, which is the maximum length of any line segment inside of the circle. All of the lengths of the rectangle would need to be less than 12. But think about it from a drawing
|dw:1433822104287:dw|
No matter what type of rectangle I draw inscribed in a circle, I can draw a diameter that will always be equal to the diagonal across that rectangle. So the length of the diagonal of the rectangle is 12, which is equal to the diameter of the circle. Can you think of a formula that solves for that diagonal?

- anonymous

umm A=2l+2w?

- anonymous

Well, the diagonal/diameter is the hypotenuse of a triangle :)

- anonymous

oh ok yea

- anonymous

i know one side is 12 i need one more side right

- anonymous

Basically, relating the rectangle to the diameter of the circle could be done through pythagorean theorem. It's one formula that can relate the dimensions of the circle and the rectangle at the same time.

- anonymous

So you know pythagorean theorem I would assume. I'll just use different variables to stay consistent with what Ive been using \(L^{2} + W^{2} = d^{2}\). And we know the value of the diameter, it's 12. So we would have \(L^{2} + W^{2} = 144\)
Following me so far?

- anonymous

ok yea

- anonymous

Right. Well that was the major goal, make a relation between the rectangle and the circle. What this does, and what we're pretty much always looking for, is gives us a formula that relates all the information possible (In as few relevant variables as possible, too).
The next step is a substitution. We want to maximize Area = LW. Well, I'm sure you've found maximums and minimums via finding critical points. Well, that's what we want to do here, but we dont want two variables L and W, we want only one. That's where the other equation we have comes into play \(L^{2} + W^{2} = 144\). We'll solve for L or W and then substitute it into the rectangular area formula we want to maximize. So let's just solve for W. This would give us:
\(W = \sqrt{144-L^{2}}\)
Now we substitute that into the area formula and we get
Area = \(L\sqrt{144-L^{2}}\)
Still okay?

- anonymous

ok

- anonymous

So do you remember how to find critical points?

- anonymous

yes take derv

- anonymous

ok so i have a question about this "insribed" probs , will they all have an triangle of some sort?

- anonymous

Well, any rectangle can be split up into 2 triangles. Thinking of it in that context is the trick to being able to come up with the 2nd formula we need for the substitution we did. I would think some sort of triangle would always be involved. I've seen problems where it's a cylinder inscribed inside of a sphere, but that still used the idea of forming some sort of triangle. So I would think so.

- anonymous

Pythagorean theorem and the idea of similar triangles come up a lot in optimization and related rates problems.

- anonymous

ok lets contiue

- anonymous

Okay, so we need the derivative of \(L\sqrt{144-L^{2}}\). Think you can get that for me?

- anonymous

ok for cv i get 2

- anonymous

cv?

- anonymous

crictcal values

- anonymous

Ah. Shouldn't have had that. What was your derivative, if you can type or draw it?

- anonymous

|dw:1433824081376:dw|

- anonymous

Yeah, looks like the 2nd end of your product rule came up wrong. So looks like the derivative of the square root portion gave ya trouble. So that derivative is a power rule followed by a chain rule. So this is what we would have:
\((1)\sqrt{144-L^{2}} + L(1/2)(144-L^{2})^{-1/2}(-2L)\)
\(= \sqrt{144-L^{2}} - L^{2}/\sqrt{144-L^{2}}\)
Can you see how I did that produce rule? Want to make sure you understand the derivative before we continue.

- anonymous

oh yea nvm i didnt write down the whole square root thing so i did it wrong

- anonymous

Ah, okay. So understand the result I got for the derivative then?

- anonymous

yep

- anonymous

ok got the answer as 8.48

- anonymous

@Concentrationalizing u got the same thing right ? could u help with more max amd mins ?

- anonymous

Okay, good job. I'd keep it as \(6\sqrt{2}\), though. Leaving it like this is an exact answer where 8.48 is a rounded answer. Unless of course the question wants a rounded figure. But yeah, using \(6\sqrt{2}\), we can plug it into the equation from before and solve for W
\(L^{2} + W^{2} = 144\)
\((6\sqrt{2})^{2}+W^{2} = 144\)
\(72 + W^{2} = 144\)
\(W = 6\sqrt{2}\)
So those would be the dimensions, both the length and width are \(6\sqrt{2}\)
As a recap, the idea of these optimization problems is to get 2 formulas. One formula represents what we want to maximize or minimize while the other formula is something that relates as much of the information given as possible in as few variables as possible. We needed to relate the rectangle and the circle and we were able to do that through pythagorean theorem.
Next is a substitution, followed by taking the derivative and finding critical values. Las is back substitution in order to get the final result :)

- anonymous

And I'm actually heading off to bed. Got school in the morning and I'm tired. I just wanted to finish this problem with ya first :P

- anonymous

ok thank you

- anonymous

you're welcome. good luck :)

- anonymous

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