total area= PIr^2+ ((9-2r)/4)^2
so find r when dA/dr=0
lengths of two pieces: 2r, 9-2r
you will get two solutions, one for min area, one for max area.
A wire 9 meters long is cut into two pieces. One piece is bent into a square for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures? Give the length of wire used for each:
For the square?
For the circle?
Where should the wire be cut to maximize the total area? Again, give the length of wire used for each:
For the square?
For the circle?
3 answers
n-7w
A wire 60 inches long is to be cut into two pieces. One of the pieces will be bent into the shape of the circle and the other into the shape of an equilateral triangle. What is the maximum sum of the areas in sq. inch of the circle and triangle possible?