Question

Let f(x,y,z):=xy+z² be a function on a unit square B:={(x,y,z)∈R³,x²+y²+z²≤1}. (a) Explain why f has a minimum and a maximum on B. (b) Show that f doesn't have a local extrema on the interior {(x,y,z)∈R³,x²+y²+z²≤1}. (c) Calculate the minimum and maximum of f with the Lagrange method on the edge {(x,y,z)∈ R³,x²+y²+z²=1}

Answer 1

The function f(x,y,z) defined over a unit sphere B has extremal values due to the compactness of B. It has no local extrema on the interior because the gradient cannot be zero within the interior. The extremal values on the sphere's surface are found using Lagrange multipliers involving the constraint g(x,y,z) = x² + y² + z² - 1.

Step-by-step explanation:

The student's question revolves around a function f(x,y,z) = xy + z2 defined over the set B, which is a unit sphere {(x,y,z) ∈ R3, x2 + y2 + z2 ≤ 1}. We first note that B is closed and bounded and therefore compact. By the extreme value theorem, f must attain a minimum and maximum on B.

For part (b), since f is differentiable, any local extremum in the interior of B would require that the gradient of f be zero. However, no such point exists within the interior where the gradient is zero. Therefore, f does not have any local extrema on the interior.

For part (c), to find the extrema on the surface of B where x2 + y2 + z2 = 1, we use Lagrange multipliers. This involves setting the gradient of f equal to a multiple of the gradient of the constraint g(x,y,z) = x2 + y2 + z2 - 1 and solving the resulting system of equations.