Optimization continued..

[legacyofpiracy]

Sorry to keep bothering all of you, but I find this forum incredibly helpful. Just seeing the problem approached in a different mannor often helps alot.

The Problem:

A rectangle is to be inscribed under the arch of the curve y=4cos(0.5x) from x=-pi to x=pi, What are the dimensions of the rectangle with the largest area, and what is the largest area?


Alright so I graphed the curve on my calculator and set the domain between negative and positive pi. I roughly sketched a rectangle and reasoned that the length would be 2x and the width f(x) or 4cos(.5x)

A=l*w
A=2x*4cos(0.5x)

A'= 2x(4-sin(0.5)*0.5)+2(4cos(0.5x)
then i set this equal to zero and used the solver program on my calculator to determine that x=-.9467

I got stuck here...wondering what to do next. Up to this point does my work appear to be correct?

 

[Unco]

G'day Legacyofpiracy,

Your work is great up to

A'= 2x(4*-sin(0.5x)*0.5)+2(4cos(0.5x)
then i set this equal to zero and used the solver program on my calculator to determine that x=-.9467

Either way, if you plug in x=-0.9467, I don't think you get zero.

I could be mistaken, but I cannot see a solution coming out without the use of numerical methods (or a computer as you have). There are only two in (-pi, pi) which are just sign opposites so you shouldn't need to check the second derivative.

Once you have that fixed (perhaps you plugged it in incorrectly?), you have x. You said the width of the rectangle was 2x and the height was 4cos(0.5x); there is no problem here.

 

[legacyofpiracy]

thanks for pointing out that error, i fixed it on the calculator and now it told me the factor was x=-.863 which doesnt seem right either. Perhaps i am missing a step, but i thought then i could substitue x into 2x and 4cos(.5x)? If this is the case the book claims the width should be approx 3.44 and the height 2.61

 

[Unco]

I agree with your book. The solutions for x in (-pi, pi) are -1.72 and 1.72.

I don't know what's happening with your calculator.

 

[legacyofpiracy]

hmm, how did you come up with that?

 

[Unco]

The Newton-Raphson method.

\(\displaystyle f(x_n) = -4x\sin{\frac{x}{2}} + 8\cos{\frac{x}{2}}\)

\(\displaystyle f'(x_n) = -2x\cos{\frac{x}{2}} - 8\sin{\frac{x}{2}}\)

\(\displaystyle x_{n+1} = x_n - \frac{-4x\sin{\frac{x}{2}} + 8\cos{\frac{x}{2}}}{-2x\cos{\frac{x}{2}} - 8\sin{\frac{x}{2}}}\)

With \(\displaystyle x_0 = \frac{\pi}{2}\), after 2 or 3 iterations the solution is quite accurate for 1.7206...

The same goes for beginning with \(\displaystyle x_0 = \frac{-\pi}{2}\) to get -1.72.

If you're not familiar with numerical methods, you might want to ask someone for help with your calculator.