A wire 17 inches long is cut into two pieces. One piece is bent to form a square and the other is bent to form a rectangle that is twice as wide as it is high. How should the wire be cut in order to minimize the total area of the square and rectangle?
2 answers
There are several similar related questions below. I'm sure you can adapt one of those solutions to your specific numbers.
4 s + 2w + 4w = 4 s + 6 w = 17
A = s^2 + w(2w) = s^2 + 2 w^2
s = (17-6w)/4
A = (1/16)(17-6w)^2 + 2 w^2
dA/dw = 0 at min
0 = (1/8)(17-6w)(-6) = 4 w
(3/4)(17-6w) = 4 w
3 (17-6w) = 16 w
51 - 18 w = 16 w
34 w = 51
w = 51/34 = 1.5
then find s and w and 2 w
A = s^2 + w(2w) = s^2 + 2 w^2
s = (17-6w)/4
A = (1/16)(17-6w)^2 + 2 w^2
dA/dw = 0 at min
0 = (1/8)(17-6w)(-6) = 4 w
(3/4)(17-6w) = 4 w
3 (17-6w) = 16 w
51 - 18 w = 16 w
34 w = 51
w = 51/34 = 1.5
then find s and w and 2 w
1 answer