Optimization of the distance from the point on an ellipse

[Saptarshi Sarkar]

Homework Statement

Find the point on an ellipse $4x^2 + 5y^2 = 20$ that is farthest away from the point (0,-2).

Relevant Equations

$D=\sqrt{x^2+(y+2)^2}$

My Attempt :We need to maximize

$D=\sqrt{x^2+(y+2)^2}$

subject to the constraint

$4x^2 + 5y^2 = 20$.

From the constraint equation, we can write

$x^2=\frac{20-5y^2}{4}$

Using this in the formula for distance,

$D=\sqrt{\frac{20-5y^2}{4}+(y+2)^2}$

Differentiating this wrt y, and equating it to 0,

$\frac{5y}2=4y+8$

Solving this, we get y = 8

But, this can't be the answer as it doesn't even lie on the ellipse. The correct answer should be (0,2) but I didn't even get 2 as a possible answer.

What did I do wrong?

PS : I know that I can use Lagrange's method. I did that and got the correct answer. But I want to know what is the mistake that I made in this one.

 

[Infrared]

You have to think about bounds: You are trying to maximize the function $D(y)=\sqrt{\frac{20-5y^2}{4}+(y+2)^2}$ on the interval $y\in [-2,2]$ (since this is the range of possible $y$ values). Since you found that $D$ has no critical points in the interior of this interval, the maximum value must be obtained at an endpoint- so you should check $y=-2$ and $y=2$ individually, and find that $D$ is maximized at $y=2$.