Question

Problem 1 You are given the maximum and minimum of the function f(x, y, z) = x2 - y2 on the surface x2 + 2y2 + 3z2 = 1 exist. Use Lagrange multiplier method to find them. Let us recall the extreme value theorem we discussed before the spring break: Extreme Value Theorem (For Functions Of Two Variables) If f(x,y) is continuous on a closed, bounded region D in the plane, then f attains a maximum value f(x,y) and a minimum value f(x2,y2) at points (21, yı) and (02, y2) in D.

Answer 1

let, Let, f(249,2)=x²-42, Let the surface S: x +2y +37=11 g(x, 92), V = x4+247372 Obne rue that the surface h a ellipsoid? which is cloned and bounded, Therefore, by en extreme value theorem f1211,7) altains the amaximum and minimum on the surface x²+2y2+37²=1. By the method of Lagranges multipliers we have to solve the good following equations. v f(44,7) = 1 og 249,7) 90944,7)=1 Therefore weget, ' Lagrange mutiplier anter 04 = 20 = 12x – 0 at an 7 -2y =197 — a is called 22+ 2y?+372=1 - ©

from ③ we have 20(1-7)=0 270 or 231 From ® we haves 2y (22+1) =0 7y zo on a = -72 From ③ we have 06720 72=0 OP Z=0

Then the solutions of , , , @ are (0,0,1), (0, taro), (+1,0,0) These are the critical point of f1247, 2), Mouy fro,o, I F 3 ) = 0 f(0, I 1 2 10) = -2 (+ , , ) = 1 Hence, the maximum value of focaly, its is í , which is attained at the 2 points (1,0,0), (-1,0,0) . The minimum values of f(241,2). s - , which is attained at Co, u 10), (0-12, 0)