題目
Problem
7. Let f ( x , y ) = 2 x 3 − 12 x y + y 3 + 13 f(x, y) = 2x^3 - 12xy + y^3 + 13 f ( x , y ) = 2 x 3 − 12 x y + y 3 + 13 P = ( p , q ) P=(p, q) P = ( p , q ) R 2 \mathbb{R}^2 R 2 f ( x , y ) f(x, y) f ( x , y ) i + j \mathbf{i}+\mathbf{j} i + j ( p , q ) = ( 11 ) ‾ . (p, q) = \underline{\quad (11) \quad} \,. ( p , q ) = ( 11 ) .
解答
解法一
思路
展開
給定方向向量 v = i + j \mathbf{v} = \mathbf{i}+\mathbf{j} v = i + j u = 1 2 ( i + j ) \mathbf{u} = \frac{1}{\sqrt{2}}(\mathbf{i}+\mathbf{j}) u = 2 1 ( i + j )
函數 f ( x , y ) f(x,y) f ( x , y ) u \mathbf{u} u D u f ( x , y ) = ∇ f ( x , y ) ⋅ u D_{\mathbf{u}}f(x, y) = \nabla f(x, y) \cdot \mathbf{u} D u f ( x , y ) = ∇ f ( x , y ) ⋅ u
計算梯度並代入公式，得到一個以 x , y x, y x , y
利用配方法求此二次函數的極小值點 ( p , q ) (p, q) ( p , q )
答題過程
展開
單位方向向量為：
u = 1 2 i + 1 2 j \mathbf{u} = \frac{1}{\sqrt{2}}\mathbf{i} + \frac{1}{\sqrt{2}}\mathbf{j} u = 2 1 i + 2 1 j 計算梯度 ∇ f ( x , y ) \nabla f(x, y) ∇ f ( x , y )
f x = 6 x 2 − 12 y , f y = − 12 x + 3 y 2 ⟹ ∇ f ( x , y ) = ⟨ 6 x 2 − 12 y , 3 y 2 − 12 x ⟩ f_x = 6x^2 - 12y, \quad f_y = -12x + 3y^2 \implies \nabla f(x, y) = \langle 6x^2 - 12y, 3y^2 - 12x \rangle f x = 6 x 2 − 12 y , f y = − 12 x + 3 y 2 ⟹ ∇ f ( x , y ) = ⟨ 6 x 2 − 12 y , 3 y 2 − 12 x ⟩ 在點 P ( p , q ) P(p,q) P ( p , q ) u \mathbf{u} u
D u f ( p , q ) = ∇ f ( p , q ) ⋅ u = 1 2 ( 6 p 2 − 12 q + 3 q 2 − 12 p ) = 1 2 ( 6 p 2 − 12 p + 3 q 2 − 12 q ) D_{\mathbf{u}}f(p, q) = \nabla f(p, q) \cdot \mathbf{u} = \frac{1}{\sqrt{2}} \left( 6p^2 - 12q + 3q^2 - 12p \right) = \frac{1}{\sqrt{2}} \left( 6p^2 - 12p + 3q^2 - 12q \right) D u f ( p , q ) = ∇ f ( p , q ) ⋅ u = 2 1 ( 6 p 2 − 12 q + 3 q 2 − 12 p ) = 2 1 ( 6 p 2 − 12 p + 3 q 2 − 12 q ) 我們欲求使此值最小的 ( p , q ) (p,q) ( p , q ) p p p q q q
6 p 2 − 12 p = 6 ( p − 1 ) 2 − 6 6p^2 - 12p = 6(p-1)^2 - 6 6 p 2 − 12 p = 6 ( p − 1 ) 2 − 6 3 q 2 − 12 q = 3 ( q − 2 ) 2 − 12 3q^2 - 12q = 3(q-2)^2 - 12 3 q 2 − 12 q = 3 ( q − 2 ) 2 − 12 故：
D u f ( p , q ) = 1 2 [ 6 ( p − 1 ) 2 + 3 ( q − 2 ) 2 − 18 ] D_{\mathbf{u}}f(p, q) = \frac{1}{\sqrt{2}} \left[ 6(p-1)^2 + 3(q-2)^2 - 18 \right] D u f ( p , q ) = 2 1 [ 6 ( p − 1 ) 2 + 3 ( q − 2 ) 2 − 18 ] 由於實數的平方項非負， 6 ( p − 1 ) 2 ≥ 0 6(p-1)^2 \ge 0 6 ( p − 1 ) 2 ≥ 0 3 ( q − 2 ) 2 ≥ 0 3(q-2)^2 \ge 0 3 ( q − 2 ) 2 ≥ 0
p − 1 = 0 ⟹ p = 1 , q − 2 = 0 ⟹ q = 2 p - 1 = 0 \implies p = 1, \quad q - 2 = 0 \implies q = 2 p − 1 = 0 ⟹ p = 1 , q − 2 = 0 ⟹ q = 2 時，方向導數取得最小值（最小變化率為 − 18 2 = − 9 2 -\frac{18}{\sqrt{2}} = -9\sqrt{2} − 2 18 = − 9 2
因此，最值點為 ( p , q ) = ( 1 , 2 ) (p, q) = \boxed{(1, 2)} ( p , q ) = ( 1 , 2 )
