Question

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

Answer 1

Set

f(x,y) = x^2 +2y^2

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

∇f(x,y)=λ∇g(x,y)∇f(x,y)=λ∇g(x,y)andg(x,y)=0.g(x,y)=0.We compute∇f(x,y)=⟨−1,2y⟩∇f(x,y)=⟨−1,2y⟩and∇g(x,y)=⟨4x+2y,2x+2y⟩.∇g(x,y)=⟨4x+2y,2x+2y⟩.The vector equality⟨−1,2y⟩ =λ⟨4x+2y,2x+2y⟩⟨−1,2y⟩ =λ⟨4x+2y,2x+2y⟩is equivalent to the coordinate-wise equalities−12y=λ(4x+2y)=λ(2x+2y).−1=λ(4x+2y)2y=λ(2x+2y).

Solving for λλ in each equation gives

λλ=−14x+2y=yx+y.λ=−14x+2yλ=yx+y.Since λλ must take on a consistent value throughout, the two right-hand sides of the above equations must be equal, and we can use this to solve for xx in terms of yy:yx+yy(4x+2y)4xy+2y24xy+xx(4y+1)x=−14x+2y=−(x+y)=−x−y=−(2y2+y)=−(2y2+y)=−2y2+y4y+1.yx+y=−14x+2yy(4x+2y)=−(x+y)4xy+2y2=−x−y4xy+x=−(2y2+y)x(4y+1)=−(2y2+y)x=−2y2+y4y+1.

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

g(x,y)2x2+2xy+y2−12(2y2+y)2(4y+1)2−22y3+y24y+1+y2−1=0=0=0.g(x,y)=02x2+2xy+y2−1=02(2y2+y)2(4y+1)2−22y3+y24y+1+y2−1=0.

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

y≈−1.38,−0.31,−0.21,1.40.y≈−1.38,−0.31,−0.21,1.40.

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):

f(0.54,−1.38)f(−0.54,−0.31)f(0.81,−0.21)f(−0.81,1.40)≈1.37≈0.63≈−0.76≈2.76.f(0.54,−1.38)≈1.37f(−0.54,−0.31)≈0.63f(0.81,−0.21)≈−0.76f(−0.81,1.40)≈2.76.

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).