a wire 34 cm long is cut into 2 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 long. How should the wire be cut in order to minimize the total area of the square and the rectangle?

Question

anonymous · Answer

We know: $$
34 = x+y\
4s = x \
2l+2w= y \
A = s^2 + lw
$$We have 5 variables and four equations, meaning we have one free variables...
I'd solve for $A$ in terms of $x$.  Then we can try to maximize $A$.

anonymous · Answer

By the way, $x$ and $y$ are the lengths of each piece.  $s$ the the side of the square made by the $x$ piece.  $l,w$ are length and width of the rectangle made from the $y$ piece.

anonymous · Answer

Seems I should have added another equation in: $$
34 = x+y\
4s = x \
2l+2w= y \
2l = w \
A = s^2 + lw
$$Sorry, so it's actually 6 variables and 5 equations.

anonymous · Answer

thank you!! after that what do i do??