Consider the function f (x, y) given by 1 f(x, y) = x2 - y2 + 5x)! and the curve C given by x = %'/1 y2 with x S 0. Use the Lagrangian multiplier method to find the maximum and minimum points of f (x, y) on the curve C. Let f(m,y) = a3. Use the method of Lagrange multipliers to find the maximum and minimum points of f(a:,y) on the curve defined by y2 l m4 :133 = 0. The maximum value is O The minimum value is l We have f(x,'y) 21: and g(a:,y) = y2 +:c4 x3 = 0. Step 1: Vf (may) : AVQ (33,291), 9033:\") : 0 We get Vf = (1,0) and V9 2 (4:33 3332, 2y), so we need to solve 1 = A (4:133 3x2) 0=Aow 0 = y2 + $4 _ :33 Clearly A 75 0 because of the first equation, so hence the second equation gives y = 0. Thus, from the third equation we have 0 = .134 :33 = :33 (a: 1). So a: = 0 or a: = 1. If a; = 0 then the first equation is not satisfied, so the only point which satises all three equations is (1,0). SteP 2: v.9 (73,351) 2 (030): 9(15 3') = 0 We have Vg(:1:,y) = (4:173 3:132, 2y) 2 (0,0) when 0 = 43:3 33:2 = 3:2 (4w 3) and when 2y 2 0. So, we get points (0, 0) and G, 0). However, (g, 0) is not on the constraint curve. Step 3: End Points The curve is a closed curve so no end points. 80, our only points are (1,0) and (0,0). We get f(1,0) = 1 is the minimum and f(0,0) = 0 is the maximum