Optimization of the distance from the point on an ellipse

  • #1
Saptarshi Sarkar
99
13
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.
 
Physics news on Phys.org
  • #2
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##.
 
Last edited:
  • Informative
Likes Saptarshi Sarkar
Back
Top