SOLUTION: A 30 cm piece of wire is cut in two. Once piece is bent into the shape of a square and the other piece bent into the shape of a rectangle with 2:1 ratio. What are the lengths of th

Algebra ->  Length-and-distance -> SOLUTION: A 30 cm piece of wire is cut in two. Once piece is bent into the shape of a square and the other piece bent into the shape of a rectangle with 2:1 ratio. What are the lengths of th      Log On


   

Question 887291: A 30 cm piece of wire is cut in two. Once piece is bent into the shape of a square and the other piece bent into the shape of a rectangle with 2:1 ratio. What are the lengths of the two pieces if the sum of the area of the square and the rectangle is a minimum?
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Two starting lengths from the wire, u%2Bv=30.

The square: x for one side length. Area is x%5E2.

The rectangle: 2y to y may be the sides ratio. Area is 2y%5E2.

If assign S as the sum of the areas, then highlight_green%28S=x%5E2%2B2y%5E2%29.

Accounting for the sum of perimeters to be 30 is also needed.
Back with u and v,
u=4x and v=2%282y%29%2B2y----based on the given ratio of its dimensions.
u=4x and v=6y.
4x%2B6y=30
highlight_green%282x%2B3y=15%29---THIS equation can be used to substitute for either x or y in the S equation. You can then determine the minimum value for S.

Maybe you can take the pathway to the solution described.