Lagrange multiplier question

[soccerfool14]

Question 5 (12 points). Let f (x, y) = x2 + y2 + x.

1. (a) Find the critical points of f in the region {(x, y) : 2x2 + 3y2 ≤ 9}.
   
2. (b) Use Lagrange multipliers to find the maxima and minima of f on the
   ellipse 2x2 + 3y2 = 9.
   
3. (c) What are the absolute maxima and minima of f on the region {(x,y) :
   2x2 + 3y2 ≤ 9}?
   
   
   
   
   This is a question on the practice final for my calc class. I understand how the points (-1/2,0), (-3/2, sqrt(3/2)),(-3/2,-sqrt(3/2)) were obtained, but the solution shows a fourth and fifth point, (+/-3/sqrt(2),0)). How was this last point obtained?!
   Thanks​

 

[stapel]

Let f(x,y) = x^2 + y^2 + x.
a) Find the critical points of f in the region {(x,y): 2x^2 + 3y^2 <= 9}.
b) Use Lagrange multipliers to find the maxima and minima of f on the ellipse 2x^2 + 3y^2 = 9.
c) What are the absolute maxima and minima of f on the region {(x,y): 2x^2 + 3y^2 <= 9}?

I understand how the points ... were obtained, but...

To which part of the exercise do "the points" apply? Thank you!

 

[HallsofIvy]

There is only one part that mentions "points"!

 

[stapel]

There is only one part that mentions "points"!

Granted. But I was assuming that the max/min points were probably "points", too, and was hoping that the original poster would reply with his work or reasoning.

 

[HallsofIvy]

Question 5 (12 points). Let f (x, y) = x2 + y2 + x.

1. (a) Find the critical points of f in the region {(x, y) : 2x2 + 3y2 ≤ 9}.
2. (b) Use Lagrange multipliers to find the maxima and minima of f on the
   ellipse 2x2 + 3y2 = 9.
3. (c) What are the absolute maxima and minima of f on the region {(x,y) :
   2x2 + 3y2 ≤ 9}?
   
   
   
   
   This is a question on the practice final for my calc class. I understand how the points (-1/2,0), (-3/2, sqrt(3/2)),(-3/2,-sqrt(3/2)) were obtained, but the solution shows a fourth and fifth point, (+/-3/sqrt(2),0)). How was this last point obtained?!
   Thanks​

Inside the ellipse, the partial derivatives of the object function, 2x+1 and 2y, are 0 at (-1/2, 0). On the ellipse, we can take \(\displaystyle x= \frac{3}{\sqrt{2}}cos(t)\), \(\displaystyle y= \sqrt{3}sin(t)\) and the object function becomes \(\displaystyle \frac{3}{2}cos^2(t)+ 3sin^2(t)+ \frac{3}{\sqrt{2}}cos(t)\). The derivative of that is \(\displaystyle -3cos(t)sin(t)+ 6sin(t)cos(t)-\frac{3}{\sqrt{2}}sin(t)= (3cos(t)- \frac{3}{\sqrt{2}}) sin(t)\). That will be 0 when sin(t)= 0 (at t= 0 and \(\displaystyle t= \pi\)) and when \(\displaystyle cos(t)= \frac{1}{\sqrt{2}}\) (at \(\displaystyle t= \frac{\pi}{4}\) and \(\displaystyle t= \frac{7\pi}{4}\).

You appear to have missed the "sin(t)= 0" solutions.