# Another optimization word prob.

• Nov 22nd 2006, 11:56 AM
Nitz456
Another optimization word prob.
Quote:

http://img75.imageshack.us/img75/615/circstt8.png
Explain why the minimum value of x^2+y^2 on x+y=4 occurs at the point at which a graph of x^2+y^2=Constant is tangent to the line x+y=4

Then

Using this answer, and implicit differentiation to find the slope of the circle, find the minimum value of x^2+y^2 such that x+y=4.
This assignment is my Everest.
• Nov 22nd 2006, 12:43 PM
CaptainBlack
Quote:

http://img75.imageshack.us/img75/615/circstt8.png
Explain why the minimum value of x^2+y^2 on x+y=4 occurs at the point at which a graph of x^2+y^2=Constant is tangent to the line x+y=4
Suppose x^2+y^2=k^2, this meets the line when:

x^2+(4-x)^2=k^2,

or:

2x^2-8x+(16-k^2)=0

which has roots:

x= [8+/-sqrt(64-8(16-k^2))]/4=2 +/- sqrt(-4+k^2/2)

Now for large k this has two solutions but as k is reduced there comes a point
where k^2/2=4, where there is only one root, any smaller and there are no roots.

So the minimum k for which there are roots corresponds to the line being
a tangent.

Quote:

Then

Using this answer, and implicit differentiation to find the slope of the circle, find the minimum value of x^2+y^2 such that x+y=4.
This seems over complex but at a point of tangency the slope of the
circle will be equal to that of the line (-1).

d/dx(x^2+y^2)=2x +2y dy/dx=0,

so:

dy/dx=-x/y.

The slope of the line is -1, so x=y at the point of tangenct, so:

2x=4, so x=2, y=2 and hence the minimum of x^2+y^2=8.

RonL
• Nov 22nd 2006, 02:35 PM
Nitz456
Thank you thank you thank you thank you