1z=xy x+y=1z=x(1-x)=-x^2+x=-(x-1/2)^2+1/4z'x=-2x+1x=1/2,z'x=0, z最大=1/42∫D∫(x^2+y^2)dxdy
x=ρcosθ,y=ρsinθx^2+y^2=ρ^2
dxdy=(1/2)(dρ)(θdθ)D x^2+y^2=1,ρ=1x^2+y^2=4 ρ=2
0<=θ<=π/2∫∫(x^2+y^2)dxdy=∫[1,2]ρ^2dρ∫[0,π/2] (1/2)θdθ=∫[1,2](π^2/4)ρ^2dρ=(7π^2/12)3(cosnπ/3)^2 =(1/2)[1+cos(2nπ/3)]-1/... |
