Quote:
Originally Posted by twisted
I need some help figuring out how to find the dimensions to this question:
A rectangle has perimeter 64cm and area 23cm^2. Solve the following system of equations to find the rectangle's dimensions.
a) l = 23/w
l + w = 32
b) solve the system of equations
x^2 + y^2 = 1
xy = 0.5
Please and thank you! I know the answer, I just don't know how to do the question.

Since Mr Hui answered part a), I'll try part b).
Since xy = 0.5 then x = 1/(2y)
Substituting for x gives:
(1/(2y))^2 + y^2 = 1
1/(4y^2) + y^2 = 1
(1 + 4y^4)/(4y^2) = 1
1 + 4y^4 = 4y^2
4y^4  4y^2 + 1 = 0
y^4  y^2 + 1/4 = 0
(y^2  1/2)(y^2  1/2) = 0
y^2  1/2 = 0
y^2 = 1/2
y = + or  sqrt(1/2)
x = + or  sqrt(1/2)