Find the max/min values of f(x,y)=2x^2+3y^2–4x–5 on the domain x^2+y^2 =100
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Find the max/min values of f(x,y)=2x^2+3y^2–4x–5 on the domain x^2+y^2 =100

[From: ] [author: ] [Date: 12-06-11] [Hit: ]
) = 299].Points for best answer, thanks for any help!!! :)-First,......
Find the maximum and minimum values of the function f(x,y)=2x^2+3y^2–4x–5 on the domain x^2+y^2 less than or equal to 100. List points lexicographically.

I've discovered that the min = -7 [ f(1, 0) = -7).
And I know that the max = 299 but but I cannot get the correct points to match [ie: f( ?, ?) = 299].

Points for best answer, thanks for any help!!! :)

-
First, we find critical points inside the domain.

f_x = 4x - 4
f_y = 6y

Setting these equal to 0 yields (x, y) = (1, 0), which is inside x^2 + y^2 = 100.
-----------------------
Now, we find points on the boundary of the circle (call it g) via Lagrange Multipliers.
grad(f) = λ grad(g)
==> <4x - 4, 6y> = λ <2x, 2y>.

Equate like entries:
4x - 4 = 2λx ==> 2x - 2 = λx
6y = 2λy ==> (λ - 3)y = 0 ==> λ = 3 or y = 0.

(i) If λ = 3, then 2x - 2 = 3x ==> x = -2.
Substituting this into x^2 + y^2 = 100 yields y = ±√96.

(ii) If y = 0, then substituting this into x^2 + y^2 = 100 yields x = ±10.

Testing all of the critical points:
f(1, 0) = -7 <----Minimum
f(-2, ±√96) = 299 <---Maximum
f(-10, 0) = 235
f(10, 0) = 155.

I hope this helps!
1
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