Answer:
Step-by-step explanation:
f(x,y)=2x2+3y2−4x−1 ⇒ ∇f =⟨4x−4,6y⟩=⟨0,0⟩ ⇒ x=1, y = 0. Thus (1,0) is the only critical point of f, and it lies in the region x2+y2<16.Ontheboundary,g(x,y)=x2+y2=16 ⇒ λ∇g=⟨2λx,2λy⟩,so6y=2λy ⇒ eithery=0orλ=3. Ify=0,then x=±4;ifλ=3,then4x−4=2λx ⇒ x=−2andy=±2√3. Now f(1,0) = −3, f(4,0) = 15, f(−4,0) = 47, and f−2,±2√3 = 51. Thus the maximum value of f(x,y) on the disk x2+y2 ≤ 16 is f−2,±2√3 = 51, and the minimum value is f (1, 0) = −3.
