# Thread: More Optimization Problems . . .

1. ## More Optimization Problems . . .

I need some help with this problem.

Find the points on the ellipse 4x^2 + y^2 = 4 that are farthest away from the point (1, 0).

d = sqrt[(x -1)^2 + (y - 0)^2]

4x^2 + y^2 = 1
=> x^2 + y^2/4 = 1
=> x^2 = 1 - y^2/4
=> x = sqrt(1 - y^2/4)
=> x = 1 - y/2 = -y

d = sqrt[(-y -1)^2 + (y - 0)^2]
d^2 = f(y) = (-y -1)^2 + (y -0)^2
f'(y) = 2(-y -1) + 2(y - 0) = -2y -2 + 2y - 0 = -2

I don't know what to do next, and I'm sure I've made a couple of mistakes anyway.

Thanks.

2. Originally Posted by zachb
I need some help with this problem.

Find the points on the ellipse 4x^2 + y^2 = 4 that are farthest away from the point (1, 0).

d = sqrt[(x -1)^2 + (y - 0)^2]
First, if you maximise d you also maximise d^2, using d^2 avoids that
sqrt so makes everything easier, so we want to maximise:

D2 = d^2 = (x-1)^2 + (y-0)^2 = (x-1)^2 + y^2

but from the equation for the ellipse:

y^2 = 4 - 4x^2

So:

D2 = (x-1)^2 - 4(x^2 -1) = - 3 x^2 - 2 x + 5

Now:

d(D2)/dx = -6 x -2

for a maximum of D2 d(D2)/dx=0, so x=-1/3, and so D2 = 16/3, and y=4sqrt(2)/3
(assuming I have done the algebra correctly).

RonL

3. You did the algebra correctly. Thanks. Solving for y^2 and them plugging it in actually made the problem a lot easier.