# Challenge I came up with: what is the maximum area of the rectangle inscribed between the line #y=sintheta#, #x=theta# and #x=pi-theta#?

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I conjured up this problem randomly, so I'm not sure if my wording is right. What is the maximum area for a rectangle inscribed in the curve #y=sinx# ? What value of #theta# would there be for this?

I conjured up this problem randomly, so I'm not sure if my wording is right. What is the maximum area for a rectangle inscribed in the curve

# "Area" ~~ 1.12219267638209 ... #

Let us set up the following variables:

# {

```
(x,"width of the rectangle"),
(y, "height of the rectangle"),
(A, "Area of the rectangle ")
```

:} #

The width of the rectangle is:

The Height of the rectangle is:

The Area of the rectangle is given by:

We cannot solve this equation analytically and so we must use Numerical Techniques, the solution gainied is:

And so,

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To find the maximum area of the rectangle inscribed between the lines (y = \sin(\theta)), (x = \theta), and (x = \pi - \theta), we can use calculus. First, let's denote the length of the rectangle as (l) and the width as (w). The area of the rectangle is (A = lw). We need to express (l) and (w) in terms of (\theta).

The length of the rectangle, (l), is the difference between the (x)-coordinates of the two vertical lines: [ l = (\pi - \theta) - \theta = \pi - 2\theta ]

The width of the rectangle, (w), is the difference between the (y)-coordinates of the horizontal line (y = \sin(\theta)) and the (x)-axis, which is (0): [ w = \sin(\theta) - 0 = \sin(\theta) ]

Now, we can express the area (A) solely in terms of (\theta): [ A(\theta) = (\pi - 2\theta) \cdot \sin(\theta) ]

To find the maximum area, we take the derivative of (A) with respect to (\theta), set it equal to zero, and solve for (\theta). Then, we check the second derivative to confirm it's a maximum.

[ A'(\theta) = \frac{dA}{d\theta} = \pi\cos(\theta) - 2\sin(\theta) - 2\theta\cos(\theta) ]

Setting (A'(\theta)) equal to zero and solving for (\theta): [ \pi\cos(\theta) - 2\sin(\theta) - 2\theta\cos(\theta) = 0 ]

Unfortunately, the exact solutions for this equation are not simple and require numerical methods to find. Once you find the critical points, you need to check the second derivative:

[ A''(\theta) = \frac{d^2A}{d\theta^2} ]

Evaluate (A''(\theta)) at each critical point. If (A''(\theta) < 0), then (\theta) corresponds to a maximum. Calculate the corresponding maximum area (A(\theta)) using the values of (\theta) found.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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