A52 inch wire is to be cut. one piece is to be bent into the shape of a square, whereas the other piece is to be bent into the shape of a rectangle whose length is twice the width. find the width of the rectangle that will minimize the total area. what is the width of the rectangle that will minimize the total area?

Let the width of the rectangle be W in.
Therefore, length = 2W in
Area of the rectangle, Ar = L*W = 2W*W = 2W^2
Also, perimeter of the rectangle, Pr = 2(L+W) = 2(2W+W) = 2(3W) = 6W
Then, perimeter of square, Ps = 52-6W
And, area of the square, As = [(52-6W)/4]^2 = [13-1.5W]^2 = (13-1.5W)(13-1.5W) = 169-19.5W-19.5W+2.25W^2 = 2.25W^2-39W+169

Therefore,
Total area, At = Ar+As = 2W^2+2.25W^2-39W+169 = 4.25W^2 -39W+169

For maximum area, the first derivative of the total area expression should be zero. Therefore;
dAt/dW = 8.5W - 39 = 0 => 8.5W = 39 => W = 39/8.5 = 4.588 in

Therefore, for maximum area, width (W) should be 4.588 in

X=y+4 is the answe