# Thread: Area of a rectangle under a cosine curve?

1. ## Area of a rectangle under a cosine curve?

What is the equation of the rectangle under the curve? The curve is 3cos(x).

2. Originally Posted by Myung

What is the equation of the rectangle under the curve? The curve is 3cos(x).
Equation of the rectangle? If you are asking what

$\displaystyle 3\int\cos(x)dx=3\sin(x)+C$

3. Is that all? My guess is 6xcos(x) because x+x times 3cos(x) with something like -pi/2 < x < pi/2 as the restrictions, is that incorrect?

4. Originally Posted by Myung
Is that all? My guess is 6xcos(x) because x+x times 3cos(x) with something like -pi/2 < x < pi/2 as the restrictions, is that incorrect?
I have no idea what you area asking, are you asking fo the area between $\displaystyle \left[\frac{-\pi}{2},\frac{\pi}{2}\right]$?

Given by $\displaystyle 3\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\cos(x)$?

5. I'm looking for an equation for the area of the rectangle inscribed under the curve 3cos(x) in the picture, the rectangle represents the dotted lines from -x to x. I don't really know where to start.

6. Originally Posted by Myung

What is the equation of the rectangle under the curve? The curve is 3cos(x).
Originally Posted by Myung
I'm looking for an equation for the area of the rectangle inscribed under the curve 3cos(x) in the picture, the rectangle represents the dotted lines from -x to x. I don't really know where to start.
The Area of the rectangle is $\displaystyle A(x)=Base*Height=6x\cos(x)$

Are we supposed to find the value of x that maximizes the area?

Thus, $\displaystyle \frac{dA}{dx}=-6x\sin(x)+6\cos(x)$

Setting this equal to zero, we get that $\displaystyle \cot(x)=x$. Using your calculator, we see that this is the case for $\displaystyle x=\pm.86 \ x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$

We take the positive value.

Thus, the area of the rectangle is $\displaystyle A(.86)=6(.86)\cos(.86)\approx \color{red}\boxed{3.37}$.

I believe this is the answer.

Hope that this makes sense to you!

--Chris

7. Originally Posted by Myung
I'm looking for an equation for the area of the rectangle inscribed under the curve 3cos(x) in the picture, the rectangle represents the dotted lines from -x to x. I don't really know where to start.
This is a maximum problem, you know the width is $\displaystyle 2x$ and the heigth is $\displaystyle 3\cos(x)$ since cosine is symetric

So you need to maximize

$\displaystyle A=6x\cos(x)$

8. Originally Posted by Chris L T521
The Area of the rectangle is $\displaystyle A(x)=Base*Height=6x\cos(x)$

Are we supposed to find the value of x that maximizes the area?

Thus, $\displaystyle \frac{dA}{dx}=-6x\sin(x)+6\cos(x)$

Setting this equal to zero, we get that $\displaystyle \cot(x)=x$. Using your calculator, we see that this is the case for $\displaystyle x=\pm.86 \ x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$

We take the positive value.

Thus, the area of the rectangle is $\displaystyle A(.86)=6(.86)\cos(.86)\approx \color{red}\boxed{3.37}$.

I believe this is the answer.

Hope that this makes sense to you!

--Chris
I think this might be it, not completely sure but i'll check in class. Thanks