Area of a rectangle under a cosine curve?

**Myung** · June 25th 2008, 04:39 PM

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

**Mathstud28** · June 25th 2008, 04:47 PM

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

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

**Myung** · June 25th 2008, 04:51 PM

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?

**Mathstud28** · June 25th 2008, 04:56 PM

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 $\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ ?

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

**Myung** · June 25th 2008, 05:00 PM

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.

**Chris L T521** · June 25th 2008, 05:10 PM

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 $A(x)=Base*Height=6x\cos(x)$

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

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

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

We take the positive value.

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

I believe this is the answer.

Hope that this makes sense to you!

--Chris

**Mathstud28** · June 25th 2008, 05:12 PM

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 $2x$ and the heigth is $3\cos(x)$ since cosine is symetric

So you need to maximize

$A=6x\cos(x)$

**Myung** · June 25th 2008, 05:18 PM

Originally Posted by Chris L T521

The Area of the rectangle is $A(x)=Base*Height=6x\cos(x)$

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

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

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

We take the positive value.

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

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

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