Math 241, Quiz 9. 10/24/11. Name:
•Read problems carefully. Show all work. No notes, calculator, or text.
•There are 15 points total.
§14.8, #5 (15 points): Use Lagrange multipliers to find the maximum and minimum values of the
function f(x, y) = x2ysubject to the constraint g(x, y) = x2+ 2y2= 6.Show your work.
Solution: We compute
∇f(x, y) = h2xy, x2i, g(x, y) = h2x, 4yi.
Setting ∇f(x, y) = λg(x, y ), we solve the system
2xy = 2λx, x2= 4λy, g(x, y ) = x2+ 2y2= 6.
First, we have
2xy = 2λx =⇒x(y−λ) = 0 =⇒x= 0 or y=λ.
1. Suppose that x= 0. Substituting in x2= 4λy yields λy = 0. Hence, we have λ= 0 or
y= 0.
(a) If y= 0, then we have g(x, y) = g(0,0) = 6 since x= 0 also, a contradiction. So
f(x, y)does not have a max or min at (0,0).
(b) If λ= 0, we only get more information from g(x, y ) = g(0, y)=6. In particular,
2y2= 6 implies that y=±√3. Therefore, f(x, y)may have a max or min at (0,±√3).
2. Suppose that y=λ. Substituting in x2= 4λy gives x2= 4y2, which implies that x=±2y.
Substituting x2= 4y2in g(x, y) = 6 gives x2+ 2y2= 4y2+ 2y2= 6y2= 6, which holds
if and only if y=±1. Since x=±2y, we find that f(x, y )may have a max or min at the
points (±2,1) and (±2,−1).
It remains to evaluate f(x, y) = x2yat the six points where f(x, y)could have a max or min. We
find that
f(0,±√3) = 0, f(±2,1) = 4, f(±2,−1) = −4.
It follows that f(x, y)has a max at (±2,1) and a min at (±2,−1).