We have ∇f(x,y,z) = ⟨4x3,4y3,4z3⟩ and ∇g(x,y,z) = ⟨2x,2y,2z⟩, so LaGrange’s method gives requires that we solve the following system of equations:
x2 + y2 + z2 = 1
We split into four cases, depending on whether x and y are zero or not:
4x3 = 2λx 4y3 = 2λy 4z3 = 2λz
(1) (2) (3) (4)
(a) x and y are both nonzero. Then equations (1) and (2) tell us that x2 = y2 = λ/2, and putting √√ √√√
this into equations (3) and (4) gives solutions (± 2/2, ± 2/2, 0) and (± 3/3, ± 3/3, ± 3/3). (b) x̸=0buty=0. Thenwehavex2 =λ/2,from(1)andputtingthisinto(4)givesλ/2+z2 =1,
√√ which using (3) gives solutions (±1, 0, 0) and (± 2/2, 0, ± 2/2).
(c) y ̸= 0 but x = 0. This is just like case (b) but with x and y reversed: the solutions are (0, ±1, 0) √√
and (0, ± 2/2, ± 2/2).
(d) x = y = 0. Then equation (4) tells us that z = ±1, so we get the two solutions (0, 0, ±1).
Now we determine which of these points are maxima and minima by simply evaluating f at all these points. We find that the maximum value of f is 1 and occurs at (±1, 0, 0), (0, ±1, 0), and (0, 0, ±1),
√√√√√√
while the minimal value is 1/3, and occurs at (± 3/3, ± 3/3, ± 3/3), (± 3/3, ± 3/3, ± 3/3), √√√
and (± 3/3, ± 3/3, ± 3/3).
