求解三道高数题，二元函数的极值，求详解

F(x,y)=x·y·(12-x-y)=12xy-x²y-xy²

∂F/∂x=12y-2xy-y²=y(12-2x-y)

∂F/∂y=12x-2xy-x²=y(12-2y-x)

驻点(4,4) (0,0)(舍去)

∂²F/∂x²=-2y

∂²F/∂x∂y=12-2x-2y

∂²F/∂x²=-2x

A=∂²F/∂x²(4,4)=-8

B=∂²F/∂x∂y(4,4)=-4

C=∂²F/∂y²(4,4)=-8

B²-AC<0,A<0→(4,4)是极大值点→三个数是4,4,4

a=½√(x²+y²+z²)→4a²=x²+y²+z²→z=√(4a²-x²-y²)

V=x·y·√(4a²-x²-y²)

∂V/∂x=y√(4a²-x²-y²)-x²y/√(4a²-x²-y²)=y(4a²-2x²-y²)/√(4a²-x²-y²)

∂V/∂y=x√(4a²-x²-y²)-xy²/√(4a²-x²-y²)=x(4a²-2y²-x²)/√(4a²-x²-y²)

驻点(2a/√3,2a/√3),经判断为极大值点→长=宽=高=2a/√3

f(x,y)=10x+9y-C=8x+6y-400-0.01(3x²+xy+3y²)

∂f/∂x=8-0.01(6x+y)

∂C/∂x=6-0.01(x+6y)

驻点：(120,80),经判断为极大值点...