æ± å½æ°f(xï¼y)=x³+y³-3xyçæå¼
解ï¼ä»¤∂f/∂x=3x²-3y=0ï¼å¾x²-y=0...........(1)
å令∂f/∂y=3y²-3x=0ï¼å¾y²-x=0...............(2)
ç±(1)å¾y=x²ï¼ä»£å
¥(2)å¼å¾x⁴-x=x(x³-1)=x(x-1)(x²+x+1)=0ï¼æ
å¾x₁=0ï¼x₂=1ï¼
ç¸åºå°ï¼y₁=0ï¼y₂=1ï¼æ
å¾é©»ç¹M(0ï¼0)åN(1ï¼1)ï¼
A=∂²f/∂x²=6xï¼B=∂²f/∂x∂y=-3ï¼C=∂²f/∂y²=6yï¼
é©»ç¹M: A=0ï¼B=-3ï¼C=0ï¼B²-AC=9>0ï¼æ
ç¹Mä¸æ¯æå¼ç¹ï¼
é©»ç¹N: A=6>0ï¼B=-3ï¼C=6ï¼B²-AC=9-36=-27<0ï¼æ
Næ¯è¯¥å½æ°çæå°ç¹ï¼
minf(xï¼y)=f(1ï¼1)=1+1-3=-1.
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