Applied Optimization

[jjhova702]

A rectangle is to be inscribed under the arch of the curve y=4cos(x/2) from x = - (pi)radians and x= (pi) radians. What are the dimensions of the rectangle with the largest area? My professor wants me to work in radians and use Newton's method to find the zero of the first derivative.
Any help would be greatly appreciated.

 

[tkhunny]

Radians: Excellent choice. How would you do it any ohter way? You DO know that radians are just numbers? It's not actually a unit, right?

Newton's Method: Okay, that should work. Do you know what it is?

1) Did you draw a picture. My quick sketch suggests we have some symmetries and the optimal rectangle is likely to be centered horizontally at the Origin. Among other thing, we need to look only on [0,pi]. Why is this?
2) Pondering the picture, the area of any inscribed rectangle would be:
----- Width: 2x
----- Height: 4cos(x/2)
3) Finish up! Let's see what you get.

Note: You must learn to ask youself questions so that you can get started yourself. What do I know? What does it want? Can I visualize it? What relationships exist? What conditions must be met?

 

[jjhova702]

Okay
A= (2x)(4 cos (x/2)) = 8xcos (x/2)


dA/dz = 8 cos (x/2) - 4xsin(x/2)

if I set dA/dx to 0. I will get

0 = 8 cos (pi/2) - 4pi Sin (pi/2) <-------------------- is this the answer?

 

[tkhunny]

Where did you get x = pi? That makes no sense. This would create a rectangle with zero height. That's not very big.

Also, it's -4pi, not 0.

Further, Newton's Method is very unlikely to result in an exact value with pi in it.

You have the derivative right, now find where it is zero. No more guessing.