Asked by Selena
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?
Answers
Answered by
Steve
There are several similar related questions below. I'm sure you can adapt one of those solutions to your specific numbers.
Answered by
Damon
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
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