Lagrange multipliers to find extreme values of the function f(x, y) = x^2 +2y^2 on the unit circle x^2 +y^2 =1

Set

so that our goal is to maximize f(x,y) = x^2 +2y^2

subject to x^2 +y^2 =1

By the method of Lagrange multipliers, we need to find simultaneous solutions to

Solving for λλ in each equation gives

We must also satisfy the constraint g(x,y)=0g(x,y)=0, so plugging this in, we need

At this point, we have reduced the problem to solving for the roots of a single variable polynomial, which any standard graphing calculator or computer algebra system can solve for us, yielding the four solutions

Plugging these back in to x=−2y2+y4y+1x=−2y2+y4y+1 gives the corresponding xx-values of approximately 0.54,−0.54,0.81,−0.810.54,−0.54,0.81,−0.81. Our four candidates for the constrained global extrema, then, are the points (0.54,−1.38),(−0.54,−0.31),(0.81,−0.21),(−0.81,1.40)(0.54,−1.38),(−0.54,−0.31),(0.81,−0.21),(−0.81,1.40). To figure out which one is the global maximum we were asked for, we simply plug them all into f(x,y)f(x,y):

We conclude that the maximum value of f(x,y)=y2−xf(x,y)=y2−x subject to the constraint g(x,y)=2x2+2xy+y2−1=0g(x,y)=2x2+2xy+y2−1=0 is 2.762.76, occurring at the point (−0.81,1.40)(−0.81,1.40).

5/20 so 25%

step-by-step explanation:

i recommend you do what your heart desires. if you're attracted to the same gender, then you're attracted to the same gender. you can't determine what your heart wants.