Calculus Question v.99999999

[Ace_O_Spades]

Find the largest possible area of a rectangle inscribed in the unit circle by maximizing as a function of the angle indicated in the figure below.

[goateebird]

Answer is 2, I think

Use A = (2sin(theta))^2 + (2cos(theta))^2, and x^2 + y^2 = 1 as your two functions. Substitute cos and sin for x and y. Combine the functions, take the derivative, set equal to zero, and go from there. It's been a while since I've done this, but I think that's right.

[KnifeMissile]

The answer is 2 but I don't think your formulas are correct. I'm not even sure what you are thinking.

The area is simply A = (2 * sin(t)) * (2 *cos(t))

Take the derivative, equate to zero, and then isolate for the variable, t.
Now, isolating for a variable after taking the derivative does not always give you the maximum! It merely gives you a local extrema, which may or may not be what you are looking for. It just so happens that, in this question, the closest extrema to t = 0 is also the maximal area...

[synic213]

The funny thing about this problem, and similar questions, is that the largest single rectangular area that can be inscribed into a circle of any size is a perfect square. In this problem the sides of the perfect square measure 2*sin(45) = 1.414..., the Area = 2. The farther from "square" the rectangle gets, the smaller the area.

[goateebird]

Originally posted by KnifeMissle
The answer is 2 but I don't think your formulas are correct. I'm not even sure what you are thinking.

The area is simply A = (2 * sin(t)) * (2 *cos(t))

Take the derivative, equate to zero, and then isolate for the variable, t.
Now, isolating for a variable after taking the derivative does not always give you the maximum! It merely gives you a local extrema, which may or may not be what you are looking for. It just so happens that, in this question, the closest extrema to t = 0 is also the maximal area...

That doesn't take into account the equation of the circle. However, I do see the mistake I made, i'm not sure why I had those exponents in the area equation. Whoops.

Regardless, the answer is 2, and I hope you get the general idea of how to do it, hehe.

[KnifeMissile]

If you think that the formula A = (2 * sin(t)) * (2 * cos(t)) "doesn't take into account the equation of a circle" then are you saying this formula is incorrect? That this function will not yield the area of the rectangle?

I'm divided as to whether I should explain this to you or let you work it out for yourself. There's something to be said for figuring out one's own mistake (I've learned a lot through this process!). On the other hand, if it doesn't occur to you I will end up coming back here and explaining it all to you, anyways...

What do the rest of you think?

I'll let you think this one over but I'll conjecture as to "why you had those exponents in the area equation." It's because you were desperately thinking you had to incorporate the cartesian equation of a circle in the area formula--somehow, anyhow...

[goateebird]

Okay, I got it. Your equation was correct, but I initally thought that it needed something to bound the rectangle from having an infinite area with no boundaries. You didn't show the implied step of substituting the cos and sin in for x and y, which I didn't notice. That substitution is what bounds it, and that's what I missed. I knew that your equation was that of the rectangle the whole time, I just missed where you bounded the equation.

Sorry for the confusion.

[Bigt6909]

calc... i just dropped calc 2... im glad i did

[Shpoop]

what binds it is not necessary the y^2+ x^2 = 1 equation...it is bound by the definition of sin and cos...they represent the unit circle, so no other equation is needed. Neither can ever be greater than 1, and they both cannot be 1 at the same time...therefore they are inherently "bounded" in this problem