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標題:

lagrange multiplier

發問:

find the highest and lowest points on the intersection of the paraboloid z=x^2+9y^2 and the cylinder x^2+y^2=1

• C.N BLUE.韓綜@1@
• 揀A定揀B定揀C-
• Lagrange Multiplier-@1@

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最佳解答:

method 1(Lagrange's multiplier) Let F(x,y,z,u,v)=z+u(x^2+9y^2-z)+v(x^2+y^2-1) (?F/?x)=2x(u+v)=0 (?F/?y)=2y(9u+v)=0 (?F/?z)=1-u x^2+9y^2-z=0 x^2+y^2-1=0 then x=0, y=+/- 1, z=9 , u=1, v=-9 or y=0, x=+/- 1, z=1, u=1, v=-1 so, max. z=9 min. z=1 method2(parameteric eq.) x^2+y^2=1, then (x,y)=(cost, sint) z=(cost)^2+9(sint)^2= 1+8(sint)^2 so, max. z= 1+8=9 min. z= 1+0=1

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