Worded problem quadratics

[PolySquared]

Hey guys,

Could someone please show me how to solve this question? Thanks.

A piece of wire 80 cm long is to be cut into two pieces. One piece is to be bent into a square and the other into a rectangle four times as long as it is wide.
a Let x cm be the length of a side of the square and y cm be the width of the rectangle. Write a formula connecting y and x.
b Let A cm^2 be the sum of the areas of the square and the rectangle.
i Find a formula for A in terms of x.
ii Find the length of both pieces of wire if A is to be a minimum.

[A TART]

So we know that total length of the wire is 80cm, hence the perimeter of the square and rectangle would =80cm

80=Perimeter of Square+Perimeter of Rectangle

Perimeter of Square= 4x (Because all four sides are equal)
Perimeter of Rectangle=4y+4y+y+y
                                     =10y (We have 2 sides which are four times as long, and the remaining 2 "y" long)

Chuck this in: 80=4x+10y

As for b, create equations that show the area of each shape (ie: x^2 for the square). You will find there will be 2 variables, one with x  and one with y. In order to remove the y, we just use the equation we found in the first step (a) by rearranging it: y=(80-4x)/10, and substituting it into the equation.

For ii), use calculus or a calculator to find the minimum of the function in a realistic domain (you cant have negative area). Once you've found x, find the perimeter of EACH OF THE SHAPES

Hope this helps.