A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Fidn the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius 6cm
anonymous
 one year ago
Fidn the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius 6cm

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you're looking for the area of a square with a diagonal length of 12. dw:1433820528039:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok how would i start , use both areas?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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
 one year ago
Best ResponseYou've already chosen the best response.0um one side of rectanlge is 12cm long ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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
 one year ago
Best ResponseYou've already chosen the best response.0Well, the diagonal/diameter is the hypotenuse of a triangle :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know one side is 12 i need one more side right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Basically, 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
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0Right. 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: \(W = \sqrt{144L^{2}}\) Now we substitute that into the area formula and we get Area = \(L\sqrt{144L^{2}}\) Still okay?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So do you remember how to find critical points?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so i have a question about this "insribed" probs , will they all have an triangle of some sort?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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
 one year ago
Best ResponseYou've already chosen the best response.0Pythagorean theorem and the idea of similar triangles come up a lot in optimization and related rates problems.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so we need the derivative of \(L\sqrt{144L^{2}}\). Think you can get that for me?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah. Shouldn't have had that. What was your derivative, if you can type or draw it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433824081376:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, 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: \((1)\sqrt{144L^{2}} + L(1/2)(144L^{2})^{1/2}(2L)\) \(= \sqrt{144L^{2}}  L^{2}/\sqrt{144L^{2}}\) Can you see how I did that produce rule? Want to make sure you understand the derivative before we continue.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh yea nvm i didnt write down the whole square root thing so i did it wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, okay. So understand the result I got for the derivative then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok got the answer as 8.48

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Concentrationalizing u got the same thing right ? could u help with more max amd mins ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, 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
 one year ago
Best ResponseYou've already chosen the best response.0And 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
 one year ago
Best ResponseYou've already chosen the best response.0you're welcome. good luck :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.